JEE Main · MathematicsHard

$If the line y = m x + c is a normal to the ellipse \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, then c^{2} is equal to:$

A. $\frac{m ^{2} ( a ^{2} - b ^{2} ) ^{2}}{a ^{2} + b ^{2} m ^{2}}$
B. $\frac{( a ^{2} - b ^{2} ) ^{2}}{a ^{2} + b ^{2} m ^{2}}$
C. $\frac{m ^{2} ( a ^{2} + b ^{2} ) ^{2}}{a ^{2} + b ^{2} m ^{2}}$
D. $\frac{( a ^{2} + b ^{2} ) ^{2}}{a ^{2} + b ^{2} m ^{2}}$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{m ^{2} ( a ^{2} - b ^{2} ) ^{2}}{a ^{2} + b ^{2} m ^{2}}$

Solution

$The equation of the normal at (a cos θ, b sin θ) is a x sec θ - b y csc θ = a^{2} - b^{2} .$
$Comparing this to y = m x + c, we identify m = (a / b) tan θ and c = - (a^{2} - b^{2}) / (b csc θ) .$
$From m, tan θ = mb / a, so sin θ = mb / a^{2} + b^{2} m^{2} .$
$Substitute sin θ into the expression for c .$
$Squaring yields c^{2} = m^{2} (a^{2} - b^{2})^{2} / (a^{2} + b^{2} m^{2}) .$
$Hence the answer is (A).$

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