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Tangents and Normals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsTangents and Normals5 Marks
Question
If the straight line $\text{x}\cos\alpha+\text{y}\sin\alpha=\text{p}$ touches the curve $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1,$ then prove that $\text{a}\cos^2\alpha-\text{b}^2\sin^2\alpha=\text{p}^2.$
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Answer

suppose the straight line $\text{x}\cos\alpha+\text{y}\sin\alpha=\text{p}$ touches the curve at $Q (x_1, y_1).$
But equation of tangent to $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}}{\text{b}^2}=1,$ at $Q (x_1, y_1)$ is
$\frac{\text{xx}_1}{\text{a}}+\frac{\text{yy}_1}{\text{b}^2}=1$
thus equation $\frac{\text{xx}_1}{\text{a}}+\frac{\text{yy}_1}{\text{b}^2}=1$ and $\text{x}\cos\alpha+\text{y}\sin\alpha=\text{p}$ represent the same line.
$\therefore\frac{\frac{\text{x}_1}{\text{a}^2}}{\cos\alpha}+\frac{\frac{\text{y}_1}{\text{b}^2}}{\sin\alpha}=\frac{1}{\text{p}}$
$\Rightarrow\text{x}_1=\frac{\text{a}^2\cos\alpha}{\text{p}},\text{y}_1=\frac{\text{b}^2\sin\alpha}{\text{p}}...(1)$
The point $Q (x_1, y_1)$ lies on the curve $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}}{\text{b}^2}=1$
$\therefore\frac{\text{a}^4\cos^2\alpha}{\text{p}^2\text{a}^2}+\frac{\text{b}^4\sin^2\alpha}{\text{p}^2\text{b}^2}=1$
$\Rightarrow\text{a}^2\cos^2\alpha-\text{b}^2\sin^2\alpha=\text{p}^2$

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