Parabola (VSAQs)

Maths-2B | 3. Parabola – VSAQs:
Welcome to VSAQs in Chapter 3: Parabola. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : Find the coordinates of the point on parabola y²=8x, whose focal distance is 10

Given parabola is $$y^2 = 8x$$

$$\Rightarrow 4a = 8$$

$$\Rightarrow a = 2$$

Given the focal distance $$SP = 10$$

Formula for the focal distance SP is $$x_1 + a$$

$$\Rightarrow x_1 + 2 = 10$$

$$\Rightarrow x_1 = 8$$

But $$y_1^2 = 8x_1$$

$$\Rightarrow y_1^2 = 8(8)$$

$$\Rightarrow y_1^2 = 64$$

$$\Rightarrow y_1 = \pm\sqrt{64}$$

$$\Rightarrow y_1 = \pm8$$

Hence $$P(x_1, y_1) = (8, \pm8)$$

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VSAQ-2 : Find the co-ordinates of the points on the parabola y²=2x whose focal distance is 5/2

Given parabola is $$y^2 = 2x$$

$$\Rightarrow 4a = 2$$

$$\Rightarrow a = \frac{1}{2}$$

Given the focal distance $$SP = \frac{5}{2}$$

The formula for the focal distance SP is $$x_1 + a$$

$$\Rightarrow x_1 + \frac{1}{2} = \frac{5}{2}$$

$$\Rightarrow x_1 = \frac{5}{2} – \frac{1}{2} = \frac{4}{2} = 2$$

Since $$y_1^2 = 2x_1$$

$$\Rightarrow y_1^2 = 2(2) = 4$$

$$\Rightarrow y_1 = \pm\sqrt{4}$$

$$\Rightarrow y_1 = \pm2$$

Therefore $$P(x_1, y_1) = (2, \pm2)$$

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VSAQ-3 : Find the value of K if the line 2y = 5x + k is a tangent to the parabola y²=6x

Given parabola is $$y^2 = 6x$$

$$\Rightarrow 4a = 6$$

$$\Rightarrow a = \frac{6}{4} = \frac{3}{2}$$

The given line is $$2y = 5x + k$$

$$\Rightarrow y = \frac{5}{2}x + \frac{k}{2}$$

Comparing with $$y = mx + c$$ we get:

$$m = \frac{5}{2}, \quad c = \frac{k}{2}$$

The tangential condition for a parabola $$y^2 = 4ax$$ is $$c = \frac{a}{m}$$

$$\Rightarrow \frac{k}{2} = \frac{3/2}{5/2}$$

$$\Rightarrow \frac{k}{2} = \frac{3}{5}$$

$$\Rightarrow k = 2 \times \frac{3}{5} = \frac{6}{5}$$

Hence, the value of k is $$\frac{6}{5}$$

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VSAQ-4 : S.T the line 2x – y + 2 = 0 is a tangent to y²=16x. Find the point of contact

Given parabola is $$y^2 = 16x$$

$$\Rightarrow 4a = 16$$

$$\Rightarrow a = 4$$

The given line is $$2x – y + 2 = 0$$

$$\Rightarrow y = 2x + 2$$

Comparing with $$y = mx + c$$ we get

$$m = 2, \quad c = 2$$

The tangential condition for a parabola $$y^2 = 4ax$$ is $$c = \frac{a}{m}$$

$$\Rightarrow \frac{a}{m} = \frac{4}{2} = 2 = c$$

Thus, the line is a tangent to the parabola.

The point of contact for the tangent $$y = mx + \frac{a}{m}$$ to the parabola $$y^2 = 4ax$$ can be given as $$\left(\frac{a}{m^2}, 2\frac{a}{m}\right)$$

$$\Rightarrow \left(\frac{4}{2^2}, 2\frac{4}{2}\right) = (1, 4)$$

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VSAQ-5 : Find the equation of tangent to the parabola y²=16x, inclined at 60° to the X-axis

Given parabola is $$y^2 = 16x$$

$$\Rightarrow 4a = 16$$

$$\Rightarrow a = 4$$

The slope of the tangent is $$m = \tan 60^\circ = \sqrt{3}$$

The equation of the tangent with slope m is given by the formula: $$y = mx + \frac{a}{m}$$

Substituting the given values:

$$y = \sqrt{3}x + \frac{4}{\sqrt{3}}$$

$$\Rightarrow y = \sqrt{3}x + \frac{4\sqrt{3}}{3}$$

$$\sqrt{3}y = 3x + 4$$

$$3x – \sqrt{3}y + 4 = 0$$

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VSAQ-6 : If (1/2,2) is one extremity of a focal chord of the parabola y²=8x. Find the coordinates of the other extremity

Given parabola is $$y^2 = 8x$$

$$\Rightarrow 4a = 8$$

$$\Rightarrow a = 2$$

Also given $$(at^2, 2at) = \left(\frac{1}{2}, 2\right)$$

$$\Rightarrow 2at = 2$$

$$\Rightarrow 2(2)t = 2$$

$$\Rightarrow 4t = 2$$

$$\Rightarrow t = \frac{1}{2}$$

If $$(at^2, 2at)$$ is one end of the focal chord then the other end is $$\left(\frac{a}{t^2}, -\frac{2a}{t}\right)$$

$$= \left(\frac{2}{\left(\frac{1}{2}\right)^2}, -2\left(\frac{2}{\frac{1}{2}}\right)\right)$$

$$= \left(2 \times 4, -4 \times 2\right)$$

$$= (8, -8)$$

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VSAQ-7 : Find the equation of the parabola whose vertex is (3,-2) focus is (3,1)

Given vertex $$A = (3,-2)$$ and Focus $$S = (3,1)$$

Here, the x-coordinates of A and S are equal, indicating the axis of the parabola is parallel to the y-axis. Also, the parabola opens vertically upward because the focus is above the vertex.

The distance $$a = AS$$ is calculated as:

$$a = \sqrt{(3-3)^2 + (1-(-2))^2} = \sqrt{0 + 9} = 3$$

The general form of a parabola with the vertex at (h,k) and axis parallel to the y-axis is:

$$(x – h)^2 = 4a(y – k)$$

$$(x – 3)^2 = 4 \times 3 \times (y + 2)$$

$$\Rightarrow (x – 3)^2 = 12(y + 2)$$

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VSAQ-8 : Find the equation of the parabola whose vertex is (1,-2), focus is (1,-7)

Given Vertex $$A = (1, -2)$$ and Focus $$S = (1, -7)$$

Here, the x-coordinates of A and S are equal, and the focus S lies below the vertex A, indicating the parabola opens vertically downward.

The distance $$a = AS$$ is calculated as:

$$a = \sqrt{(1-1)^2 + (-2-(-7))^2} = \sqrt{0 + 25} = 5$$

Also, Vertex $$A(h, k) = (1, -2)$$

The general form of a parabola with the vertex at (h,k) and axis parallel to the y-axis, opening downwards, is:

$$(x – h)^2 = -4a(y – k)$$

$$(x – 1)^2 = -4 \times 5 \times (y + 2)$$

$$\Rightarrow (x – 1)^2 = -20(y + 2)$$

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VSAQ-9 : Find the vertex, focus, equation of the directrix and axis of the parabola y²=16x

Given parabola $$y^2 = 16x$$

$$\Rightarrow 4a = 16$$

$$\Rightarrow a = \frac{16}{4} = 4$$

(i) The vertex A of the parabola is at the origin, $$A = (0,0)$$

(ii) The focus S of the parabola, given by $$S = (a, 0) S = (4,0)$$

(iii) The equation of the directrix of the parabola, which is $$x = -a x = -4$$ or equivalently, $$x + 4 = 0$$

(iv) The axis of the parabola, which is parallel to the y-axis for this standard form, is $$y = 0$$

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VSAQ-10 : Find the equation of the tangent and normal at the positive end of L.R on the parabola y²=6x

Given parabola $$y^2 = 6x$$

$$\Rightarrow 4a = 6$$

$$\Rightarrow 2a = 3$$

$$\Rightarrow a = \frac{3}{2}$$

The positive end of the Latus Rectum (L.R) is $$(a,2a) = \left(\frac{3}{2}, 3\right)$$

The equation of the tangent at $$(x_1,y_1)$$ on $$S = y^2 – 4ax = 0$$ is $$S_1 = yy_1 – 2a(x+x_1) = 0$$

$$\Rightarrow y(3) – 3(x + \frac{3}{2}) = 0$$

$$\Rightarrow 3[y – (x + \frac{3}{2})] = 0$$

$$\Rightarrow y – (x + \frac{3}{2}) = 0$$

$$\Rightarrow y = x + \frac{3}{2}$$

The slope of the tangent $$2x – 2y + 3 = 0$$

$$\Rightarrow \text{the slope of its normal is } -1$$

The equation of the normal at $$(\frac{3}{2}, 3)$$ with slope −1 is,

$$y – 3 = -1(x – \frac{3}{2})$$

$$\Rightarrow y = -x + 3 + \frac{3}{2}$$

$$\Rightarrow y = -x + \frac{9}{2}$$

$$x + y – \frac{9}{2} = 0$$

$$\Rightarrow 2x + 2y – 9 = 0$$

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VSAQ-11 : Prove that the point on the parabola y²=4ax,(a>0) nearest to the focus is its vertex

Focal distance of $$P(x_1, y_1)$$ on the parabola is $$SP = x_1 + a$$

The value of $$x_1 + a$$ is minimum when $$x_1 = 0$$

If $$x_1 = 0$$ then $$y_1^2 = 4ax_1 \Rightarrow y_1^2 = 4a(0) \Rightarrow y_1^2 = 0 \Rightarrow y_1 = 0$$

$$x_1 = 0 y_1 = 0$$

⇒ vertex A (0,0) is the nearest point to the focus.