If the line lx+my=1 is a normal to the hyperbola x2a2−y2b2=1 then a2l2−b2m2=

xx1a2−yy1b2−1=0Slope of normal=−a2y1b2x1=−1my−y1=−a2y1b2x1(x−x1)⇒ya2y1−1a2=−xb2x1+1b2⇒xb2x1+ya2y1=1a2+1b2⇒lx+my=11a2+1b2=ya2y1m=1b2x1l=k(say)⇒y1=1a2km,x1=1b2kl1b4k2l2a2−1a4k2m2b2=1a2l2−b2m2=a4b4k2⇒a4b4(1a2+1b2)2=(a2+b2)2

If the line lx+my=1 is a normal to the hyperbola x2a2−y2b2=1 then a2l2−b2m2=

xx1a2−yy1b2−1=0Slope of normal=−a2y1b2x1=−1my−y1=−a2y1b2x1(x−x1)⇒ya2y1−1a2=−xb2x1+1b2⇒xb2x1+ya2y1=1a2+1b2⇒lx+my=11a2+1b2=ya2y1m=1b2x1l=k(say)⇒y1=1a2km,x1=1b2kl1b4k2l2a2−1a4k2m2b2=1a2l2−b2m2=a4b4k2⇒a4b4(1a2+1b2)2=(a2+b2)2

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If the line $l x + m y = 1$ is a normal to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ then $\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} =$

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$a^{2} - b^{2}$

$a^{2} + b^{2}$

${(a^{2} + b^{2})}^{2}$

${(a^{2} - b^{2})}^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{x x_{1}}{a^{2}} - \frac{y y_{1}}{b^{2}} - 1 = 0 \\ Slope of normal = \frac{- a^{2} y_{1}}{b^{2} x_{1}} = \frac{- 1}{m} \\ y - y_{1} = \frac{- a^{2} y_{1}}{b^{2} x_{1}} (x - x_{1}) \\ \Rightarrow \frac{y}{a^{2} y_{1}} - \frac{1}{a^{2}} = \frac{- x}{b^{2} x_{1}} + \frac{1}{b^{2}} \\ \Rightarrow \frac{x}{b^{2} x_{1}} + \frac{y}{a^{2} y_{1}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \Rightarrow l x + m y = 1 \\ \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{y}{a^{2} y_{1} m} = \frac{1}{b^{2} x_{1} l} = k (say) \\ \Rightarrow y_{1} = \frac{1}{a^{2} k m}, x_{1} = \frac{1}{b^{2} k l} \\ \frac{1}{b^{4} k^{2} l^{2} a^{2}} - \frac{1}{a^{4} k^{2} m^{2} b^{2}} = 1 \\ \frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = a^{4} b^{4} k^{2} \Rightarrow a^{4} b^{4} {(\frac{1}{a^{2}} + \frac{1}{b^{2}})}^{2} = {(a^{2} + b^{2})}^{2} \end{array}$