Ellipse (SAQs)
Maths-2B | 4. Ellipse – SAQs:
Welcome to SAQs in Chapter 4: Ellipse. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.
SAQ-1 : Find the eccentricity, coordinates of foci, length of latus rectum and equations of directrices of the ellipse 9x2+16y2=144
Given the equation of the ellipse is $$9x^2 + 16y^2 = 144$$
$$\Rightarrow \frac{9x^2}{144} + \frac{16y^2}{144} = 1$$
$$\Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$
Here $$a^2 = 16 b^2 = 9$$ which implies $$a > b$$ Hence, the ellipse is horizontal.
(i) The eccentricity e is given by $$e = \sqrt{\frac{a^2 – b^2}{a^2}} = \sqrt{\frac{16 – 9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$
(ii) The foci are located at $$(\pm ae, 0)$$ which gives $$(\pm 4 \left( \frac{\sqrt{7}}{4} \right), 0) = (\pm \sqrt{7}, 0)$$
(iii) The length of the latus rectum is given by $$2 \frac{b^2}{a} = 2 \frac{9}{4} = \frac{9}{2}$$
(iv) The equation of the directrices is $$x = \pm \frac{a}{e} = \pm \frac{4}{\frac{\sqrt{7}}{4}} = \pm \frac{16}{\sqrt{7}}$$ which simplifies to
$$\sqrt{7}x = \pm 16$$
$$\Rightarrow \sqrt{7}x \pm 16 = 0$$
SAQ-2 : Find the eccentricity, coordinates of foci, length of latus rectum and equations of directrices of the ellipse 9x2+16y2-36x+32y-92=0
Given ellipse is $$9x^2 + 16y^2 – 36x + 32y – 92 = 0$$
$$\Rightarrow (9x^2 – 36x) + (16y^2 + 32y) = 92$$
$$\Rightarrow 9(x^2 – 4x + 4) + 16(y^2 + 2y + 1) = 92 + 36 + 16$$
$$\Rightarrow 9(x – 2)^2 + 16(y + 1)^2 = 144$$
$$\Rightarrow \frac{9(x – 2)^2}{144} + \frac{16(y + 1)^2}{144} = 1$$
$$\Rightarrow \frac{(x – 2)^2}{16} + \frac{(y + 1)^2}{9} = 1$$
Comparing with $$\frac{(x – h)^2}{a^2} + \frac{(y – k)^2}{b^2} = 1$$ we get $$a^2 = 16 b^2 = 9$$
$$\Rightarrow a = 4, b = 3$$
$$\Rightarrow a > b$$
Hence the ellipse is horizontal. Also, center $$(h,k) = (2,-1)$$
(i) Eccentricity e is given by $$e = \sqrt{\frac{a^2 – b^2}{a^2}} = \sqrt{\frac{16 – 9}{16}} = \frac{\sqrt{7}}{4}$$
(ii) Foci coordinates are $$(h \pm ae, k) = (2 \pm 4\left(\frac{\sqrt{7}}{4}\right),-1) = (2 \pm \sqrt{7},-1)$$
(iii) Length of the latus rectum is $$2\frac{b^2}{a} = 2\frac{9}{4} = \frac{9}{2}$$
(iv) Equations of the directrices are $$x = h \pm \frac{a}{e} = 2 \pm \frac{4}{\sqrt{7}/4} = 2 \pm \frac{16}{\sqrt{7}}$$
$$\Rightarrow x = 2 \pm \frac{16}{\sqrt{7}}$$
$$\sqrt{7}x = 2\sqrt{7} \pm 16$$
SAQ-3 : Find the eccentricity, foci and directrices of the ellipse 4x2+y2-8x+2y+1=0
Given the ellipse equation $$4x^2 + y^2 – 8x + 2y + 1 = 0$$
$$\Rightarrow 4x^2 + y^2 – 8x + 2y = -1$$
Rearranging and completing the squares,
$$\Rightarrow 4(x^2 – 2x + 1) + (y^2 + 2y + 1) = -1 + 4 + 1$$
$$\Rightarrow 4(x – 1)^2 + (y + 1)^2 = 4$$
$$\Rightarrow \frac{(x – 1)^2}{1} + \frac{(y + 1)^2}{4} = 1$$
Comparing this with the general form $$\frac{(x – h)^2}{a^2} + \frac{(y – k)^2}{b^2} = 1$$ we find
$$a^2 = 1 \Rightarrow a = 1, \quad b^2 = 4 \Rightarrow b = 2$$
Since a < b the ellipse is vertical. The center of the ellipse is $$(h, k) = (1, -1)$$
(i) The eccentricity e is given by: $$e = \sqrt{\frac{b^2 – a^2}{b^2}} = \sqrt{\frac{4 – 1}{4}} = \frac{\sqrt{3}}{2}$$
(ii) The foci coordinates are $$(h, k \pm be)$$
$$= (1, -1 \pm 2 \left( \frac{\sqrt{3}}{2} \right))$$
$$= (1, -1 \pm \sqrt{3})$$
(iii) The equations of the directrices, given by $$y = k \pm \frac{b}{e}$$ are
$$y = -1 \pm \frac{2}{\sqrt{3}/2}$$
$$= -1 \pm \frac{4}{\sqrt{3}}$$
$$\sqrt{3}y = -\sqrt{3} \pm 4$$
$$\sqrt{3}y + \sqrt{3} \pm 4 = 0$$
SAQ-4 : Find the eccentricity, coordinates of foci of the ellipse 3x2+y2-6x-2y-5=0
Given the ellipse equation $$3x^2 + y^2 – 6x – 2y – 5 = 0$$
$$\Rightarrow 3x^2 + y^2 – 6x – 2y = 5$$
Rearranging and completing the squares,
$$\Rightarrow 3(x^2 – 2x + 1) + (y^2 – 2y + 1) = 5 + 3 + 1$$
$$\Rightarrow 3(x – 1)^2 + (y – 1)^2 = 9$$
$$\Rightarrow \frac{(x – 1)^2}{3} + \frac{(y – 1)^2}{9} = 1$$
Comparing this with the general ellipse form $$\frac{(x – h)^2}{a^2} + \frac{(y – k)^2}{b^2} = 1$$ we identify:
$$a^2 = 3 \Rightarrow a = \sqrt{3}, \quad b^2 = 9 \Rightarrow b = 3$$
Since a < b the ellipse is vertical. The center $$(h, k) = (1, 1)$$
(i) The eccentricity e is given by: $$e = \sqrt{\frac{b^2 – a^2}{b^2}} = \sqrt{\frac{9 – 3}{9}} = \frac{\sqrt{6}}{3} = \frac{\sqrt{2}}{3}$$
(ii) The foci coordinates are $$(h, k \pm be)$$
$$= (1, 1 \pm 3\left(\frac{\sqrt{2}}{3}\right))$$
$$= (1, 1 \pm \sqrt{6})$$
SAQ-5 : Find the equations of the tangents to the ellipse 2x2+y2=8 which are
i.Parallel to x – 2y – 4 = 0
ii.Perpendicular to x + y + 2 = 0
iii.Make an angle 45° with x-axis
Given ellipse is $$2x^2 + y^2 = 8$$
$$\Rightarrow \frac{2x^2}{8} + \frac{y^2}{8} = 1$$
$$\Rightarrow \frac{x^2}{4} + \frac{y^2}{8} = 1$$
$$\Rightarrow a^2 = 4, b^2 = 8$$
(i) Slope of the line $$x – 2y – 4 = 0$$ is $$m = \frac{1}{2}$$
Equation of the tangent with slope m is $$y = mx \pm \sqrt{a^2m^2 + b^2}$$
$$\Rightarrow y = \frac{1}{2}x \pm \sqrt{4 \cdot \frac{1}{4} + 8} = \frac{1}{2}x \pm \sqrt{9} = \frac{1}{2}x \pm 3$$
$$\Rightarrow 2y = x \pm 6$$
$$\Rightarrow x – 2y \pm 6 = 0$$
(ii) Slope of the line $$x + y + 2 = 0$$ is -1
⇒ Slope of its perpendicular is $$m = 1$$
Equation of the tangent with slope m = 1 is $$y = mx \pm \sqrt{a^2m^2 + b^2}$$
$$\Rightarrow y = x \pm \sqrt{4(1)^2 + 8} = x \pm \sqrt{12} = x \pm 2\sqrt{3}$$
$$\Rightarrow x – y \pm 2\sqrt{3} = 0$$
(iii) Given that $$\theta = 45^\circ$$
$$\Rightarrow \text{Slope } m = \tan 45^\circ = 1$$
Equation of the tangent is $$y = mx \pm \sqrt{a^2m^2 + b^2}$$
$$\Rightarrow y = x \pm \sqrt{4(1)^2 + 8} = x \pm \sqrt{12} = x \pm 2\sqrt{3}$$
SAQ-6 : Find the equations of the tangents to 9×2+16y2=144, which make equal intercepts on the coordinate axes
Given ellipse is $$9x^2 + 16y^2 = 144$$
$$\Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$
$$\Rightarrow a^2 = 16 \Rightarrow a = 4$$ and $$b^2 = 9 \Rightarrow b = 3$$
The equation of any line which makes equal intercepts on the axes is given by $$x \pm y + k = 0$$ Its slope $$m = \pm 1$$
The equation of the tangent with slope m to an ellipse is $$y = mx \pm \sqrt{a^2m^2 + b^2}$$
Substituting the given values into this formula, we get:
$$y = \pm x \pm \sqrt{16(1)^2 + 9} = \pm x \pm \sqrt{16 + 9} = \pm x \pm \sqrt{25} = \pm x \pm 5$$
SAQ-7 : Find the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is 15/2
The distance between the foci $$S(ae,0)$$ and $$S'(-ae,0)$$ is 2
$$\Rightarrow 2ae = 2$$
$$\Rightarrow ae = 1 \quad \dots (1)$$
The length of the latus rectum is $$\frac{15}{2}$$
$$\Rightarrow \frac{2b^2}{a} = \frac{15}{2}$$
$$\Rightarrow b^2 = \frac{15}{4a} \quad \dots (2)$$
Now, $$b^2 = a^2(1-e^2) = a^2 – a^2e^2 = a^2 – (ae)^2 = a^2 – 1 \quad \dots (3)$$
From (2) & (3), $$\frac{15}{4a} = a^2 – 1$$
$$\Rightarrow 15a = 4a^2 – 4$$
$$\Rightarrow 4a^2 – 15a – 4 = 0$$
$$\Rightarrow (a – 4)(4a + 1) = 0$$
$$\Rightarrow a = 4 \quad \text{or} \quad -\frac{1}{4}$$
If a = 4 then $$b^2 = a^2 – 1 = 16 – 1 = 15$$
The equation of the ellipse is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$\Rightarrow \frac{x^2}{16} + \frac{y^2}{15} = 1$$
SAQ-8 : Find the equation of the ellipse in the standard form such that the distance between the foci is 8 and the distance between directrices is 32
The distance between the foci S = (ae,0) and S’ = (-ae,0) is 8
$$\Rightarrow 2ae = 8$$
$$\Rightarrow ae = 4 \quad \dots (1)$$
The distance between directrices is 32.
$$\Rightarrow 2\left(\frac{a}{e}\right) = 32$$
$$\Rightarrow \frac{a}{e} = 16 \quad \dots (2)$$
Multiplying (1) & (2) we get $$(ae)\left(\frac{a}{e}\right) = 4(16) = 64$$
$$\Rightarrow a^2 = 64$$
Now $$b^2 = a^2(1-e^2) = a^2 – a^2e^2 = a^2 – (ae)^2 = 64 – 16 = 48$$
The equation of the ellipse is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$\Rightarrow \frac{x^2}{64} + \frac{y^2}{48} = 1$$
SAQ-9 : Find the equation of the ellipse referred to its major, minor axes as the coordinate axes X.Y-respectively with latus rectum of length 4 and distance between foci 4√2
Let the equation of the ellipse be $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where a > b
Length of latus rectum is 4
$$\Rightarrow 2b^2 = 4a$$
$$\Rightarrow b^2 = 2a$$
Distance between foci S = (ae,0) and S’ = (-ae,0) is $$4\sqrt{2}$$
$$\Rightarrow 2ae = 4\sqrt{2}$$
$$\Rightarrow ae = 2\sqrt{2}$$
$$\Rightarrow (ae)^2 = (2\sqrt{2})^2 = 8$$
Now $$b^2 = a^2(1-e)^2$$
$$\Rightarrow 2a = a^2 – (ae)^2 = a^2 – 8$$
$$\Rightarrow a^2 – 2a – 8 = 0$$
$$\Rightarrow (a – 4)(a + 2) = 0$$
$$\Rightarrow a = 4$$
$$b^2 = 2a = 2(4) = 8$$
The equation of the ellipse is $$\frac{x^2}{16} + \frac{y^2}{8} = 1$$
$$\Rightarrow x^2 + 2y^2 = 16$$
SAQ-10 : Find the equation of ellipse, if focus = (1,-1), e = 2/3 and directrix is x + y + 2 = 0
Given that the focus S = (1, -1) eccentricity $$e = \frac{2}{3}$$ and the directrix is $$x + y + 2 = 0$$
Let $$P(x_1, y_1)$$ be any point on the ellipse.
$$\Rightarrow SP = e \cdot PM$$
$$\Rightarrow \sqrt{(x_1 – 1)^2 + (y_1 + 1)^2} = \frac{2}{3} \left| \frac{x_1 + y_1 + 2}{\sqrt{2}} \right|$$
$$\Rightarrow 9[(x_1 – 1)^2 + (y_1 + 1)^2] = 2(x_1 + y_1 + 2)^2$$
$$\Rightarrow 9[x_1^2 – 2x_1 + 1 + y_1^2 + 2y_1 + 1] = 2[x_1^2 + y_1^2 + 4 + 2x_1y_1 + 4y_1 + 4x_1]$$
$$\Rightarrow 9(x_1^2 + y_1^2 – 2x_1 + 2y_1 + 2) = 2(x_1^2 + y_1^2 + 2x_1y_1 + 4x_1 + 4y_1 + 4)$$
$$\Rightarrow 9x_1^2 + 9y_1^2 – 18x_1 + 18y_1 + 18 = 2x_1^2 + 2y_1^2 + 4x_1y_1 + 8x_1 + 8y_1 + 8$$
$$\Rightarrow 7x_1^2 – 4x_1y_1 + 7y_1^2 – 26x_1 + 10y_1 + 10 = 0$$
SAQ-11 : Find the equation of the ellipse whose focus is (1,2), eccentricity is 2/3 & directrix is 2x + 3y + 6 = 0
By the focus-directrix property, the equation of the ellipse with focus $$P(x, y)$$ such that
$$\frac{SP}{PM} = e$$
$$\Rightarrow SP = e \cdot PM$$
$$\Rightarrow \sqrt{(x-1)^2 + (y-2)^2} = \frac{2}{3} \left| \frac{2x + 3y + 6}{\sqrt{4 + 9}} \right|$$
Squaring and cross multiplying, we get
$$17 \left(x^2 – 2x + 1 + y^2 – 4y + 4 \right) = 4 \left(4x^2 + 9y^2 + 36 + 12xy + 36y + 24x \right)$$
$$\Rightarrow 101x^2 – 48xy + 81y^2 – 330x – 612y + 441 = 0$$
The above equation represents the equation of an oblique ellipse
SAQ-12 : Find the equation of ellipse in the standard form, if passes through the points (-2,2) and (3,-1)
The equation of the ellipse in the standard form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
If (A) passes through (-2, 2) then $$\frac{(-2)^2}{a^2} + \frac{2^2}{b^2} = 1$$
$$\Rightarrow \frac{4}{a^2} + \frac{4}{b^2} = 1…(1)$$
If (A) passes through (3, -1) then $$\frac{3^2}{a^2} + \frac{(-1)^2}{b^2} = 1$$
$$\Rightarrow \frac{9}{a^2} + \frac{1}{b^2} = 1…(2)$$
From (1) and (2) $$\frac{4}{a^2} + \frac{4}{b^2} = \frac{9}{a^2} + \frac{1}{b^2}$$
$$\Rightarrow \frac{4}{b^2} – \frac{1}{b^2} = \frac{9}{a^2} – \frac{4}{a^2}$$
$$\Rightarrow \frac{3}{b^2} = \frac{5}{a^2}$$
$$\Rightarrow \frac{1}{b^2} = \frac{5}{3a^2}$$
Now using (2):
$$\Rightarrow \frac{9}{a^2} + \frac{5}{3a^2} = 1$$
$$\Rightarrow \frac{27+5}{3a^2} = 1$$
$$\Rightarrow \frac{32}{3a^2} = 1$$
$$\Rightarrow \frac{1}{a^2} = \frac{3}{32}$$
Again using (2): $$\Rightarrow \frac{1}{b^2} = 1 – \frac{9}{a^2} = 1 – 9\left(\frac{3}{32}\right) = \frac{32-27}{32} = \frac{5}{32}$$
The equation of the ellipse is $$\frac{3}{32}x^2 + \frac{5}{32}y^2 = 1$$
$$\Rightarrow 32x^2 + 5y^2 = 32$$
SAQ-13 : Find the equation of tangent and normal to the ellipse x2+8y2-33=0 at (-1,2)
The equation of the ellipse is $$S = x^2 + 8y^2 – 33 = 0$$
$$\Rightarrow \text{The equation of the tangent at } (-1,2) \text{ is } S_1 = 0$$
$$\Rightarrow x_1x + 8y_1y – 33 = 0$$
$$\Rightarrow (-1)x + 8(2)y – 33 = 0$$
$$\Rightarrow -x + 16y – 33 = 0$$
$$\Rightarrow x – 16y + 33 = 0$$
The slope of the tangent is $$\frac{1}{16}$$
$$\Rightarrow \text{The slope of its normal is } -16$$
Equation of the normal at (-1,2) with slope -16 is:
$$y – 2 = -16(x + 1)$$
$$\Rightarrow y – 2 = -16x – 16$$
$$\Rightarrow 16x + y + 14 = 0$$
SAQ-14 : Find the equation of tangent and normal to the ellipse 9x2+16y2=144 at the end of the latus rectum in the first quadrant
Given Ellipse $$9x^2 + 16y^2 = 144$$
$$\Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$
$$\Rightarrow a^2 = 16, \, b^2 = 9$$
$$\Rightarrow a = 4, \, b = 3$$
$$e = \sqrt{\frac{a^2 – b^2}{a^2}} = \sqrt{\frac{16 – 9}{16}} = \sqrt{\frac{7}{16}}$$
$$\Rightarrow e = \frac{\sqrt{7}}{4}$$
Positive end of the latus rectum $$L = (ae, \frac{b^2}{a}) = \left(4 \cdot \frac{\sqrt{7}}{4}, \frac{9}{4}\right) = (\sqrt{7}, \frac{9}{4})$$
Equation of the tangent at L is $$S_1 = 0$$
$$\Rightarrow \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$
$$\Rightarrow \frac{x\sqrt{7}}{16} + \frac{y}{9} \cdot \frac{9}{4} = 1$$
$$\Rightarrow \sqrt{7}x + 4y = 16$$
Equation of the normal at L is $$a^2x/x_1 – b^2y/y_1 = a^2 – b^2$$
$$\Rightarrow \frac{16x}{\sqrt{7}} – \frac{9y}{9/4} = 16 – 9$$
$$\Rightarrow \frac{16x}{\sqrt{7}} – 4y = 7$$
$$\Rightarrow 16x – 4\sqrt{7}y = 7\sqrt{7}$$
SAQ-15 : Show that the points of intersection of the perpendicular tangents to an ellipse lie on a circle
Let the equation of the ellipse be $$S = \frac{x^2}{a^2} + \frac{y^2}{b^2} – 1 = 0$$
Let $$P(x_1,y_1)$$ be a point on the locus.
A tangent with slope m is given by $$y = 3x \pm \sqrt{a^2m^2 + b^2}$$
If it passes through $$P(x_1,y_1)$$ be a point on the locus.
A tangent with slope m is given by $$y = 3x \pm \sqrt{a^2m^2 + b^2}$$
If it passes through $$P(x_1,y_1)$$ then $$y_1 = mx_1 \pm \sqrt{a^2m^2 + b^2}$$
$$\Rightarrow y_1 – mx_1 = \pm \sqrt{a^2m^2 + b^2}$$
$$\Rightarrow (y_1 – mx_1)^2 = a^2m^2 + b^2$$
$$\Rightarrow y_1^2 + m^2x_1^2 – 2x_1y_1m – a^2m^2 – b^2 = 0 \quad \text{(1)}$$
Equation (1) is a quadratic equation in m and its roots are taken as m1,m2 (slopes of tangents).
If the tangents are perpendicular, then $$m_1m_2 = -1$$
From (1), the product of roots $$= \frac{y_1^2 – b^2}{x_1^2 – a^2} = -1$$
$$\Rightarrow y_1^2 – b^2 = -(x_1^2 – a^2) = -x_1^2 + a^2$$
$$\Rightarrow x_1^2 + y_1^2 = a^2 + b^2$$
The equation of the locus of $$P(x_1,y_1)$$ is $$x^2 + y^2 = a^2 + b^2$$
SAQ-16 : Show that the locus of the feet of the perpendiculars drawn from either of the foci to any tangent to the ellipse is the auxiliary circle
Let the equation of the ellipse be $$S = \frac{x^2}{a^2} + \frac{y^2}{b^2} – 1 = 0$$
Let $$P(x_1,y_1)$$ be a point on the locus.
A tangent with slope m is $$y = mx \pm \sqrt{a^2m^2 + b^2}$$
$$\Rightarrow y – mx = \pm \sqrt{a^2m^2 + b^2} \quad …(1)$$
The equation of any line perpendicular to the above tangent, with slope −1/m, and if it passes through the foci (±ae,0), then its equation is
$$y – 0 = -\frac{1}{m}(x \pm ae)$$
$$\Rightarrow my + x = \pm ae \quad …(2)$$
But $$P(x_1,y_1)$$ is the point of intersection of equations (1) and (2),
$$\Rightarrow y_1 – mx_1 = \pm \sqrt{a^2m^2 + b^2} \quad \text{and} \quad my_1 + x_1 = \pm ae$$
Squaring and adding the above two equations, we get
$$(y_1 – mx_1)^2 + (my_1 + x_1)^2 = (a^2m^2 + b^2) + (ae)^2$$
$$\Rightarrow (y_1^2 + m^2x_1^2 – 2mx_1y_1) + (m^2y_1^2 + x_1^2 + 2mx_1y_1) = a^2m^2 + b^2 + a^2e^2$$
$$\Rightarrow x_1^2(1+m^2) + y_1^2(1+m^2) = a^2m^2 + [a^2(1-e^2)] + a^2e^2$$
Since $$b^2 = a^2(1-e^2)$$ the equation simplifies to
$$(x_1^2+y_1^2)(1+m^2) = a^2(m^2+1)$$
$$\Rightarrow x_1^2 + y_1^2 = a^2$$
The locus of $$P(x_1,y_1)$$ is $$x^2 + y^2 = a^2$$
SAQ-17 : Find the condition for the line lx + my + n = 0 to be a tangent to the ellipse x2/a2 +y2/b2 =1
Let lx + my + n = 0 be the tangent at $$p(\theta) = (a\cos\theta, b\sin\theta)$$
The equation of the tangent at $$p(\theta)$$ is $$\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1$$
Comparing the above equation with lx + my = -n we get
$$\frac{\cos\theta}{al} = \frac{\sin\theta}{bm} = -\frac{1}{n}$$
$$\Rightarrow \cos\theta = -\frac{al}{n}, \sin\theta = -\frac{bm}{n}$$
Now, using the identity $$\cos^2\theta + \sin^2\theta = 1$$ we have
$$\Rightarrow \left(\frac{a^2l^2}{n^2}\right) + \left(\frac{b^2m^2}{n^2}\right) = 1$$
$$\Rightarrow a^2l^2 + b^2m^2 = n^2$$
SAQ-18 : If a tangent to the ellipse x2/a2+y2/b2=1(a>b) meets its major axis and minor axis at M and N respectively then prove that a2/(CM)2+b2/(CN)2=1 where C is the centre of the ellipse
The parametric point on the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ is $$P(\theta) = (a\cos\theta, b\sin\theta)$$
Then the equation of the tangent at p(θ) is
$$\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1$$
$$\Rightarrow \frac{x}{a/\cos\theta} + \frac{y}{b/\sin\theta} = 1$$
he tangent at P meets the major axis (X-axis) and minor axis (Y-axis) at M and N respectively.
$$\Rightarrow \text{X-intercept} = CM = \frac{a}{\cos\theta} \text{ and Y-intercept} = CN = \frac{b}{\sin\theta}$$
$$\Rightarrow \cos\theta = \frac{a}{CM} \text{ and } \sin\theta = \frac{b}{CN}$$
Now, using the identity $$\cos^2\theta + \sin^2\theta = 1$$ we have
$$\Rightarrow \frac{a^2}{(CM)^2} + \frac{b^2}{(CN)^2} = 1$$
SAQ-19 : If the normal at one end of a latus rectum of the ellipse x2/a2+y2/b2=1 with eccentricity e, passes through one end of the minor axis, then show that e4+e2=1
The equation of the normal at $$(x_1, y_1)$$ is $$a^2x/x_1 – b^2y/y_1 = a^2 – b^2$$
Let $$L = (ae, b^2/a)$$ be one end of the latus rectum
Hence, the equation of the normal at L is $$a^2x/ae – b^2y/(b^2/a) = a^2 – b^2$$
$$\Rightarrow ax/e – ay = a^2 – b^2 \quad …(1)$$
But equation (1) passes through the one end B'(0, -b) of the minor axis.
$$\Rightarrow a(0)/e – a(-b) = a^2 – b^2$$
$$\Rightarrow ab = a^2 – a^2(1-e^2)$$
$$\Rightarrow ab = a^2e^2$$
$$\Rightarrow e^2 = b/a$$
$$e^4 = \frac{b^2}{a^2} = \frac{a^2(1-e^2)}{a^2} = 1-e^2$$
$$\Rightarrow e^4 + e^2 = 1$$
SAQ-20 : S and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the eccentricity of the ellipse
Let the equation of the ellipse be $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where a > b
The two foci are S(ae,0) T = (-ae,0) and B(0,b) be one end of the minor axis
The distance between the two foci is $$ST = 2ae$$
Now $$\Delta STB$$ is equilateral
$$\Rightarrow SB = ST = TB$$
Now SB = ST
$$\Rightarrow (SB)^2 = (ST)^2$$
$$\Rightarrow (ae)^2 + b^2 = (2ae)^2 = 4a^2e^2$$
$$\Rightarrow a^2e^2 + b^2 = 4a^2e^2$$
$$\Rightarrow a^2e^2 + a^2(1-e^2) = 4a^2e^2$$
$$\Rightarrow e^2 = \frac{1}{4}$$
$$\Rightarrow e = \frac{1}{2}$$
The eccentricity of the ellipse is $$e = \frac{1}{2}$$