11 Most FAQ’s of Pair of Straight Lines Chapter in Inter 1st Year Maths-1B (TS/AP)

7 Marks

LAQ-1 : If the equation ax² + 2hxy + by² = 0 represents a pair of lines then prove that the equation of the pair of angular bisectors is h(x² – y²) – (a – b)xy = 0

Let $$ax^2 + 2hxy + by^2 = (x \sin \alpha – y \cos \alpha) (x \sin \beta – y \cos \beta)$$

On equating like term coeff we get

$$a = \sin \alpha \sin \beta b = \cos \alpha \cos \beta 2h = -\sin(\alpha + \beta)$$

Equation of angular bisectors is

$$\frac{x \sin \alpha – y \cos \alpha}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = \pm \frac{x \sin \beta – y \cos \beta}{\sqrt{\sin^2 \beta + \cos^2 \beta}}$$

$$x \sin \alpha – y \cos \alpha = \pm(x \sin \beta – y \cos \beta)$$

$$(x \sin \alpha – y \cos \alpha)^2 = (x \sin \beta – y \cos \beta)^2$$

$$x^2 \sin^2 \alpha + y^2 \cos^2 \alpha – xy(\sin 2\alpha) = x^2 \sin^2 \beta + y^2 \cos^2 \beta – 2xy \sin \beta \cos \beta$$

$$x^2 \sin^2 \alpha + y^2 \cos^2 \alpha – xy(\sin 2\alpha) = x^2 \sin^2 \beta + y^2 \cos^2 \beta – xy(\sin 2\beta)$$

$$x^2(\sin^2 \alpha – \sin^2 \beta) + y^2(\cos^2 \alpha – \cos^2 \beta) = xy(\sin 2\alpha – \sin 2\beta)$$

$$x^2(\sin^2 \alpha – \sin^2 \beta) + y^2[(1 – \sin^2 \alpha) – (1 – \sin^2 \beta)] = xy(\sin 2\alpha – \sin 2\beta)$$

$$x^2(\sin^2 \alpha – \sin^2 \beta) – y^2(\sin^2 \alpha – \sin^2 \beta) = xy(\sin 2\alpha – \sin 2\beta)$$

$$(\sin^2 \alpha – \sin^2 \beta)(x^2 – y^2) = 2 \cos(2\alpha + 2\beta/2) \sin(2\alpha – 2\beta/2)$$

$$[ \sin(\alpha + \beta) \sin(\alpha – \beta)](x^2 – y^2) = 2 \cos(\alpha + \beta) \sin(\alpha – \beta) xy$$

$$\sin(\alpha + \beta)(x^2 – y^2) = 2 \cos(\alpha + \beta)(xy)$$

$$\sin(\alpha + \beta)(x^2 – y^2) = 2(\cos \alpha \cos \beta – \sin \alpha \sin \beta) xy$$

$$-2h(x^2 – y^2) = 2(b – a) xy$$

$$h(x^2 – y^2) = (a – b) xy$$

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LAQ-2 : Prove that the area of the triangle formed by the pair of lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is n²√h² – ab/|am² – 2hlm + bl²

Given the quadratic form:

$$ax^2 + 2hxy + by^2 = (m_1 x – y)(m_2 x – y)$$

By equating the coefficients, we obtain:

$$m_1 + m_2 = – \frac{2h}{b}, \quad m_1m_2 = \frac{a}{b} \quad \text{(1)}$$

For solving the system of equations lx + my + n = 0 and $$m_1x – y = 0$$ we get point A as follows:

$$\begin{align} lx + my + n &= 0 \ m_1x – y &= 0 \end{align}$$

$$\frac{x}{m_0} – \frac{n}{-1} = \frac{y}{nm_1} – \frac{l}{0} = \frac{1}{l(-1)} – mm_1$$

$$\Rightarrow x/n = y/nm_1 = 1/-l – mm_1$$

$$\Rightarrow A = \left(-\frac{n}{l} + mm_1, -\frac{nm_1}{l} + mm_1\right)$$

$$B = \left(-\frac{n}{l} + mm_2, -\frac{nm_2}{l} + mm_2\right)$$

The area of the triangle with vertices $$O(0,0), A(x_1, y_1), B(x_2, y_2)$$ is given by:

$$\Delta = \frac{1}{2} |x_1y_2 – x_2y_1|$$

Calculating the area $$\Delta OAB$$

$$\text{Area of } \Delta OAB = \frac{1}{2} \left| \left(-\frac{n}{l} + mm_1\right)\left(-\frac{nm_2}{l} + mm_2\right) – \left(-\frac{n}{l} + mm_2\right)\left(-\frac{nm_1}{l} + mm_1\right) \right|$$

$$= \frac{1}{2} \left| \frac{n^2m_2 – n^2m_1}{l^2 + lmm_2 + lmm_1 + m_2m_1m_2} \right|$$

$$= \frac{1}{2} n^2 \sqrt{\left(- \frac{2h}{b}\right)^2 – 4\frac{a}{b}} \bigg/ \left| l^2 + lm\left(-\frac{2h}{b}\right) + m_2\left(\frac{a}{b}\right) \right|$$

$$= \frac{1}{2} n^2 \sqrt{\frac{4h^2 – 4ab}{b^2}} \bigg/ \left| bl^2 – 2hlm + am_2 \right|$$

$$= \frac{n^2 \sqrt{h^2 – ab}}{|am_2 – 2hlm + bl^2|} \text{ sq. units}$$

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LAQ-3 : P.T the product of the perpendiculars from (α,β) to ax² + 2hxy + by² = 0 is |aα² + 2hαβ + bβ²|/√(a – b)² + 4h²

Let $$ax^2 + 2hxy + by^2 = (l_1x + m_1y)(l_2x + m_2y)$$

On equating like term coefficients, we get:

$$a = l_1l_2$$

$$b = m_1m_2$$

$$2h = l_1m_2 + l_2m_1$$

The perpendicular distance from $$(\alpha,\beta)$$ to $$l_1x + m_1y = 0$$ is $$P_1 = \frac{|l_1\alpha + m_1\beta|}{\sqrt{l_1^2 + m_1^2}}$$

The perpendicular distance from $$(\alpha,\beta)$$ to $$l_2x + m_2y = 0$$ is $$P_2 = \frac{|l_2\alpha + m_2\beta|}{\sqrt{l_2^2 + m_2^2}}$$

The product of the perpendicular distances is:

$$p_1 \cdot p_2 = \left(\frac{|l_1\alpha + m_1\beta|}{\sqrt{l_1^2 + m_1^2}}\right) \left(\frac{|l_2\alpha + m_2\beta|}{\sqrt{l_2^2 + m_2^2}}\right)$$

$$= \frac{|(l_1\alpha + m_1\beta)(l_2\alpha + m_2\beta)|}{(\sqrt{l_1^2 + m_1^2})(\sqrt{l_2^2 + m_2^2})}$$

$$= \frac{|l_1l_2\alpha^2 + l_1m_2\alpha\beta + l_2m_1\alpha\beta + m_1m_2\beta^2|}{\sqrt{l_1^2l_2^2 + m_1^2m_2^2 + l_1^2m_2^2 + l_2^2m_1^2}}$$

$$= \frac{|l_1l_2\alpha^2 + (l_1m_2 + l_2m_1)\alpha\beta + m_1m_2\beta^2|}{\sqrt{(l_1l_2 – m_1m_2)^2 + 2l_1l_2m_1m_2 + (l_1m_2 + l_2m_1)^2 – 2l_1m_2l_2m_1}}$$

$$= \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{(a – b)^2 + 4h^2}}$$

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LAQ-4 : If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents two parallel lines then prove that (i) h² = ab (ii) af² = bg² (iii) the distance between the parallel lines is 2√g² – ac/a(a + b) (or) 2√f² – bc/b(a + b)

Given the equation:

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 = (lx + my + n_1)(lx + my + n_2)$$

On equating like term coefficients, we get:

$$a = l^2 b = m^2 h = lm 2g = l(n_1 + n_2) 2f = m(n_1 + n_2) c = n_1n_2$$

Further derivations:

$$h^2 = (lm)^2 = l^2m^2 = ab$$

$$\Rightarrow h^2 = ab$$

$$af^2 = l^2 \left(\frac{m(n_1 + n_2)}{2}\right)^2 = l^2m^2\frac{(n_1 + n_2)^2}{4} = m^2l^2\frac{(n_1 + n_2)^2}{4} = m^2\left(\frac{l(n_1 + n_2)}{2}\right)^2 = bg^2$$

Distance between $$lx + my + n_1 = 0 and lx + my + n_2 = 0$$

$$\frac{|n_1 – n_2|}{\sqrt{l^2 + m^2}} = \sqrt{(n_1 + n_2)^2 – 4n_1n_2} \Big/ \sqrt{a + b} = \sqrt{\left(\frac{2g}{l}\right)^2 – 4c} \Big/ \sqrt{a + b} = \sqrt{\frac{4g^2}{l^2} – \frac{4c}{a + b}} = \sqrt{\frac{4g^2 – 4ac}{a(a + b)}} = \frac{2\sqrt{g^2 – ac}}{a(a + b)}$$

Similarly, by taking $$n_1 + n_2 = \frac{2f}{m}$$ we get the distance between the lines as:

$$\frac{2\sqrt{f^2 – bc}}{b(a + b)}$$

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LAQ-5 : If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines then prove that (a) ∆ = abc + 2fgh – af² – bg² – ch² = 0 (b) h² ≥ ab, f² ≥ bc, g² ≥ ac

Let $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = (l_1x + m_1y + n_1)(l_2x + m_2y + n_2)$$

On equating like term coefficients, we find:

$$a = l_1l_2$$

$$b = m_1m_2$$

$$c = n_1n_2$$

$$2h = l_1m_2 + l_2m_1$$

$$2g = l_1n_2 + l_2n_1$$

$$2f = m_1n_2 + m_2n_1$$

$$(2h)(2g)(2f) = (l_1m_2 + l_2m_1)(l_1n_2 + l_2n_1)(m_1n_2 + m_2n_1)$$

On multiplying and regrouping, it yields:

$$8fgh = l_1l_2(m_1^2n_2^2 + m_2^2n_1^2) + m_1m_2(n_1^2l_2^2 + n_2^2l_1^2) + n_1n_2(l_1^2m_2^2 + l_2^2m_1^2) + 2l_1l_2m_1m_2n_1n_2$$

$$= l_1l_2[(m_1n_2 + m_2n_1)^2 – 2m_1m_2n_1n_2] + m_1m_2[(n_1l_2 + n_2l_1)^2 – 2n_1n_2l_1l_2] + n_1n_2[(l_1m_2 + l_2m_1)^2 – 2l_1l_2m_1m_2)] + 2l_1l_2m_1m_2n_1n_2$$

$$8fgh = a((2f)^2 – 2bc) + b((2g)^2 – 2ac) + c((2h)^2 – 2ab) + 2abc = 4af^2 + 4bg^2 + 4ch^2 – 4abc$$

$$2fgh = af^2 + bg^2 + ch^2 – abc$$

$$abc + 2fgh – af^2 – bg^2 – ch^2 = 0$$

$$h^2 – ab = \left(\frac{l_1m_2 + l_2m_1}{2}\right)^2 – l_1l_2m_1m_2$$

$$= \frac{(l_1m_2 + l_2m_1)^2 – 4l_1l_2m_1m_2}{4} = \frac{(l_1m_2 – l_2m_1)^2}{4} \geq 0$$

$$h^2 \geq ab$$

Similarly, it can be shown:

$$g^2 \geq ac$$

$$f^2 \geq bc$$

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LAQ-6 : Prove that the equation 3x² + 7xy + 2y² + 5x + 5y + 2 = 0 represents a pair of straight lines and find the coordinates of the point of intersection

(a) Given equation is $$3x^2 + 7xy + 2y^2 + 5x + 5y + 2 = 0$$

Comparing with $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ we identify:

$$a = 3 2h = 7$$

$$h = \frac{7}{2}$$

$$b = 2 2g = 5$$

$$g = \frac{5}{2}$$

$$2f = 5$$

$$f = \frac{5}{2} c = 2$$

Calculation for the determinant

$$\Delta = abc + 2fgh – af^2 – bg^2 – ch^2$$

$$= 3 \cdot 2 \cdot 2 + 2 \cdot \frac{5}{2} \cdot \frac{5}{2} \cdot \frac{7}{2} – 3 \cdot \left(\frac{5}{2}\right)^2 – 2 \cdot \left(\frac{5}{2}\right)^2 – 2 \cdot \left(\frac{7}{2}\right)^2$$

$$= 12 + \frac{175}{4} – \frac{75}{4} – \frac{50}{4} – \frac{98}{4}$$

$$= 12 + 175 – 75 – 50 – 98/4 = 0/4 = 0$$

Other conditions:

$$h^2 – ab = \left(\frac{7}{2}\right)^2 – 3 \cdot 2 = \frac{49}{4} – 6 = \frac{49 – 24}{4} = \frac{25}{4} > 0$$

$$f^2 – bc = \left(\frac{5}{2}\right)^2 – 2 \cdot 2 = \frac{25}{4} – 4 = \frac{25 – 16}{4} = \frac{9}{4} > 0$$

$$g^2 – ac = \left(\frac{5}{2}\right)^2 – 3 \cdot 2 = \frac{25}{4} – 6 = \frac{25 – 24}{4} = \frac{1}{4} > 0$$

These confirmations lead to the conclusion that the given equation represents a pair of straight lines.

(b) Calculating the point of intersection $$P = \left(\frac{hf – bg}{ab – h^2}, \frac{gh – af}{ab – h^2}\right)$$

$$hf – bg = \frac{7}{2} \cdot \frac{5}{2} – 2 \cdot \frac{5}{2} = \frac{35}{4} – \frac{10}{2}$$

$$ab – h^2 = 3 \cdot 2 – \left(\frac{7}{2}\right)^2 = 6 – \frac{49}{4} = \frac{24 – 49}{4} = -\frac{25}{4}$$

$$gh – af = \frac{5}{2} \cdot \frac{7}{2} – 3 \cdot \frac{5}{2} = \frac{35}{4} – \frac{15}{2}$$

$$\frac{gh – af}{ab – h^2} = \frac{\frac{35}{4} – \frac{15}{2}}{-\frac{25}{4}} = -\frac{1}{5}$$

$$\frac{hf – bg}{ab – h^2} = \frac{\frac{35}{4} – \frac{10}{2}}{-\frac{25}{4}} = -\frac{3}{5}$$

Thus, the point of intersection P is $$(- \frac{3}{5}, – \frac{1}{5})$$

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LAQ-7 : Find the angle between the lines joining the origin to the points of intersection of the curve x² + 2xy + y² + 2x + 2y – 5 = 0 and the line 3x – y + 1 = 0

Given line equation:

$$3x – y + 1 = 0$$

$$\Rightarrow 3x – y = -1$$

$$\Rightarrow \frac{3x – y}{-1} = 1 \quad \text{(1)}$$

Given curve equation:

$$x^2 + 2xy + y^2 + 2x + 2y – 5 = 0 \quad \text{(2)}$$

Homogenizing (1) with (2), we substitute:

$$x^2 + 2xy + y^2 + 2x\left(\frac{3x – y}{-1}\right) + 2y\left(\frac{3x – y}{-1}\right) – 5\left(\frac{3x – y}{1}\right)^2 = 0$$

$$\Rightarrow x^2 + 2xy + y^2 – 2x(3x – y) – 2y(3x – y) – 5(3x – y)^2 = 0$$

$$\Rightarrow x^2 + 2xy + y^2 – 6x^2 + 2xy – 6xy + 2y^2 – 5(9x^2 + y^2 – 6xy) = 0$$

$$\Rightarrow x^2 + 2xy + y^2 – 6x^2 + 2xy – 6xy + 2y^2 – 45x^2 – 5y^2 + 30xy = 0$$

$$\Rightarrow 50x^2 + 2y^2 – 28xy = 0$$

$$\Rightarrow 2(25x^2 + y^2 – 14xy) = 0$$

$$\Rightarrow 25x^2 + y^2 – 14xy = 0$$

From this homogenized equation, comparing with $$ax^2 + by^2 + 2hxy = 0$$

$$a = 25 b = 1 2h = -14$$

To find the angle θ between the lines:

$$\cos \theta = \frac{|a + b|}{\sqrt{(a – b)^2 + (2h)^2}}$$

$$= \frac{|25 + 1|}{\sqrt{(25 – 1)^2 + (-14)^2}} = \frac{26}{\sqrt{576 + 196}} = \frac{26}{\sqrt{772}} = \frac{26}{\sqrt{4 \times 193}} = \frac{26}{2\sqrt{193}} = \frac{13}{\sqrt{193}}$$

$$\cos \theta = \frac{13}{\sqrt{193}}$$

$$\Rightarrow \theta = \cos^{-1}\left(\frac{13}{\sqrt{193}}\right)$$

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LAQ-8 : Show that the lines joining the origin with the points of intersection of the curve 7x² – 4xy + 8y² + 2x – 4y – 8 = 0 with the line 3x – y = 2 are mutually perpendicular

Given the line:

$$3x – y = 2$$

$$\Rightarrow \frac{3x – y}{2} = 1 \quad \text{(1)}$$

Given the curve:

$$7x^2 – 4xy + 8y^2 + 2x – 4y – 8 = 0 \quad \text{(2)}$$

Homogenizing (1) with (2), you substitute the value from (1) into (2):

$$7x^2 – 4xy + 8y^2 + 2x\left(\frac{3x – y}{2}\right) – 4y\left(\frac{3x – y}{2}\right) – 8\left(\frac{3x – y}{2}\right)^2 = 0$$

$$\Rightarrow 7x^2 – 4xy + 8y^2 + x(3x – y) – 2y(3x – y) – 2(9x^2 + y^2 – 6xy) = 0$$

$$\Rightarrow 7x^2 – 4xy + 8y^2 + 3x^2 – xy – 6xy + 2y^2 – 18x^2 – 2y^2 + 12xy = 0$$

$$\Rightarrow -8x^2 + 8y^2 + xy = 0$$

From the final equation:

$$-8x^2 + 8y^2 + xy = 0$$

Here, the coefficients of x² and y² are equal in magnitude but opposite in sign (−8 and +8), and you have derived that:

$$\text{Coefficient of } x^2 + \text{Coefficient of } y^2 = -8 + 8 = 0$$

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LAQ-9 : Show that the lines joining the origin to the points of intersection of the curve x² – xy + y² + 3x + 3y – 2 = 0 and the straight line x – y – √2 = 0 are mutually perpendicular

Given the line:

$$x – y = \sqrt{2}$$

$$\Rightarrow \frac{x – y}{\sqrt{2}} = 1 \quad \text{(1)}$$

Given the curve:

$$x^2 – xy + y^2 + 3x + 3y – 2 = 0 \quad \text{(2)}$$

Homogenizing (1) with (2), substituting the expression from (1) into (2):

$$x^2 – xy + y^2 + 3x\left(\frac{x – y}{\sqrt{2}}\right) + 3y\left(\frac{x – y}{\sqrt{2}}\right) – 2\left(\frac{x – y}{\sqrt{2}}\right)^2 = 0$$

This simplifies to:

$$x^2 – xy + y^2 + \frac{3x(x – y)}{\sqrt{2}} + \frac{3y(x – y)}{\sqrt{2}} – \frac{2(x – y)^2}{2} = 0$$

$$\Rightarrow \sqrt{2}x^2 – \sqrt{2}xy + \sqrt{2}y^2 + 3x(x – y) + 3y(x – y) – \sqrt{2}(x^2 + y^2 – 2xy) = 0$$

$$\Rightarrow \sqrt{2}x^2 – \sqrt{2}xy + \sqrt{2}y^2 + 3x^2 – 3xy + 3yx – 3y^2 – \sqrt{2}x^2 – \sqrt{2}y^2 + 2\sqrt{2}xy = 0$$

$$\Rightarrow 3x^2 – 3y^2 + \sqrt{2}xy = 0$$

From the final equation:

$$3x^2 – 3y^2 + \sqrt{2}xy = 0$$

Here, the coefficients of x² and y² are equal in magnitude but opposite in sign (3 and -3):

$$\text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 3 – 3 = 0$$

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LAQ-10 : Find the value of k if the lines joining the origin with the points of intersection of the curve 2x² – 2xy + 3y² + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular

Given the line:

$$x + 2y = k$$

$$\Rightarrow \frac{x + 2y}{k} = 1 \quad \text{(1)}$$

Given the curve:

$$2x^2 – 2xy + 3y^2 + 2x – y – 1 = 0 \quad \text{(2)}$$

Homogenizing (1) with (2) involves substituting x + 2y from (1) into (2):

$$2x^2 – 2xy + 3y^2 + 2x\left(\frac{x + 2y}{k}\right) – y\left(\frac{x + 2y}{k}\right) – \left(\frac{x + 2y}{k}\right)^2 = 0$$

Expanding and simplifying:

$$2x^2 – 2xy + 3y^2 + \frac{2x(x + 2y)}{k} – \frac{y(x + 2y)}{k} – \frac{(x^2 + 4xy + 4y^2)}{k^2} = 0$$

$$\Rightarrow k^2(2x^2 – 2xy + 3y^2) + k(2x^2 + 4xy) – k(xy + 2y^2) – (x^2 + 4xy + 4y^2) = 0$$

Gathering like terms:

$$k^2(2x^2 – 2xy + 3y^2) + k(2x^2 + 4xy – xy – 2y^2) – (x^2 + 4xy + 4y^2) = 0$$

$$\Rightarrow x^2(2k^2 + 2k – 1) + y^2(3k^2 – 2k – 4) + xy(-2k^2 + 3k – 4) = 0$$

To determine if the lines represented by this quadratic form are perpendicular:

$$\text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 0$$

$$(2k^2 + 2k – 1) + (3k^2 – 2k – 4) = 0$$

$$5k^2 – 5 = 0$$

$$5(k^2 – 1) = 0$$

$$k^2 – 1 = 0$$

$$k^2 = 1$$

$$k = \pm 1$$

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LAQ-11 : Find the condition for the lines joining to the points of intersection of the circle x² + y² = a² and the line lx + my = 1 to coincide

Given the line:

$$lx + my = 1 \quad \text{(1)}$$

Given the circle:

$$x^2 + y^2 = a^2$$

$$\Rightarrow x^2 + y^2 – a^2 = 0 \quad \text{(2)}$$

Homogenizing (1) with (2):

$$x^2 + y^2 – a^2\left(\frac{lx + my}{1}\right)^2 = 0$$

$$\Rightarrow x^2 + y^2 – a^2(lx + my)^2 = 0$$

Expanding and rearranging terms:

$$x^2 + y^2 – a^2(l^2x^2 + m^2y^2 + 2lmxy) = 0$$

$$\Rightarrow x^2 + y^2 – a^2l^2x^2 – a^2m^2y^2 – 2a^2lmxy = 0$$

Grouping like terms gives:

$$x^2(1 – a^2l^2) + y^2(1 – a^2m^2) + xy(-2a^2lm) = 0$$

The condition for coincident lines (lines that overlap perfectly) is given by the relationship $$h^2 = ab$$ in the general quadratic form $$ax^2 + 2hxy + by^2 = 0$$

$$a = 1 – a^2l^2$$

$$b = 1 – a^2m^2$$

$$2h = -2a^2lm$$

Applying $$h^2 = ab$$ to check for coincident lines:

$$(-a^2lm)^2 = (1 – a^2l^2)(1 – a^2m^2)$$

$$\Rightarrow a^4l^2m^2 = 1 – a^2l^2 – a^2m^2 + a^4l^2m^2$$

Rearranging to solve for l and m:

$$a^2l^2 + a^2m^2 = 1$$