Question

The line $lx + my + n = 0$ intersects the curve $a{x^2} + 2hxy + b{y^2} = 1$ at P and Q the circle with PA as diameter passes through the origin then $\dfrac{{{l^2} + {m^2}}}{{{n^2}}}$
A. $a + b$
B. ${(a + b)^2}$
C. ${a^2} + {b^2}$
D. ${a^2} - {b^2}$

Answer 1

Hint : In this question, we are given the equation of a line and a curve. The point of intersection of these two will satisfy the equation of both of them. So, we will simplify one of the equations such that one of the variables is expressed in terms of the other. Then putting this value in the other equation, we will find some relations between the points P and Q. Now, a circle passes through P, Q and origin and PQ is the diameter so using the properties of the circle, we can find out the correct answer.

Complete step by step solution:
We are given that the line intersects the curve at P and Q, let the coordinates of P be $({x_1},{y_1})$ and the coordinates of Q be $({x_2},{y_2})$ .
So, we get –
$lx + my + n = 0 \\\ \Rightarrow y = \dfrac{{ - n - lx}}{m} \;$
On putting this value in the second equation, we get –
$a{x^2} + 2hx(\dfrac{{ - n - lx}}{m}) + b{(\dfrac{{ - n - lx}}{m})^2} = 1 \\\ \Rightarrow a{x^2} + \dfrac{{ - 2hnx - 2hl{x^2}}}{m} + \dfrac{{b{n^2} + b{l^2}{x^2} + 2bnlx}}{{{m^2}}} = \\\ \Rightarrow a{m^2}{x^2} - 2hnmx - 2hlm{x^2} + b{n^2} + b{l^2}{x^2} + 2bnlx = {m^2} \\\ \Rightarrow (a{m^2} - 2hlm + b{l^2}){x^2} + (2bnl - 2hmn)x + b{n^2} - {m^2} = 0 \;$
The above equation is a quadratic equation in terms of x, we know that for a quadratic equation $e{x^2} + fx + g = 0$ , the product of its roots is equal to $\dfrac{e}{g}$ . In the above equation, we have $e = a{m^2} - 2hlm + b{l^2}$ and $g = b{n^2} - {m^2}$ , let two roots of the above equation be ${x_1}$ and ${x_2}$
So, we have ${x_1}{x_2} = \dfrac{{b{n^2} - {m^2}}}{{a{m^2} - 2hlm + b{l^2}}}$
We can also write –
$lx + my + n = 0 \\\ \Rightarrow x = \dfrac{{ - n - my}}{l} \;$
On putting this value in the second given equation, we get –
$a{(\dfrac{{ - n - my}}{l})^2} + 2h(\dfrac{{ - n - my}}{l})y + b{y^2} = 1 \\\ \Rightarrow \dfrac{{a{n^2} + a{m^2}{y^2} + 2anmy}}{{{l^2}}} + \dfrac{{ - 2hny - 2hm{y^2}}}{l} + b{y^2} = 1 \\\ \Rightarrow a{n^2} + a{m^2}{y^2} + 2anmy - 2hnly - 2hlm{y^2} + b{l^2}{y^2} = {l^2} \\\ \Rightarrow (a{m^2} - 2hlm + b{l^2}){y^2} + (2amn - 2hnl)y + a{n^2} - {l^2} = 0 \;$
We get –
${y_1}{y_2} = \dfrac{{a{n^2} - {l^2}}}{{a{m^2} - 2hlm + b{l^2}}}$
Now, we know that PQ is the diameter and the circle passes through the origin O, The triangle formed by the two corners of the diameter and any point lying on the circle is a right-angled triangle with right angle at the point lying on the circle, so OP is perpendicular to OQ. So, the product of their slopes will be equal to -1.
$\Rightarrow \dfrac{{{y_1} - 0}}{{{x_1} - 0}} \times \dfrac{{{y_2} - 0}}{{{x_2} - 0}} = - 1 \\\ \Rightarrow \dfrac{{{y_1}{y_2}}}{{{x_1}{x_2}}} = - 1 \\\ \Rightarrow {y_1}{y_2} = - {x_1}{x_2} \;$
On putting the known values in the above equation, we get –

$$\dfrac{{a{n^2} - {l^2}}}{{a{m^2} - 2hlm + b{l^2}}} = - (\dfrac{{b{n^2} - {m^2}}}{{a{m^2} - 2hlm + b{l^2}}}) \\\ \Rightarrow a{n^2} - {l^2} = - b{n^2} + {m^2} \\\ \Rightarrow (a + b){n^2} = {m^2} + {l^2} \\\ \Rightarrow \dfrac{{{l^2} + {m^2}}}{{{n^2}}} = a + b \;$$

So, the correct answer is “Option A”.

Note : A polynomial equation is said to be quadratic when the degree of the equation is 2, the solutions of an equation are defined as those values of the unknown variable at which the value of the polynomial equation comes out to be zero. The roots obey certain properties like the one shown in the above solution.