Q98 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics
If a normal drawn to the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ touches the hyperbola $\frac{x^2}{4}-\frac{y^2}{3}=1$, then the square of the slope of that normal is
- Correct answerA$\frac{1+\sqrt{17}}{4}$
- B$\frac{-1+\sqrt{17}}{4}$
- C$\frac{-1+\sqrt{37}}{4}$
- D$\frac{1+\sqrt{37}}{4}$
Solution
For normal of ellipse:
$\frac{x^2}{4}+\frac{y^2}{3}=1$ ...(i)
$y^{\prime}=-\frac{3}{4} \frac{x}{y}$
Slope of normal is $m=-\frac{1}{y^{\prime}}=\frac{4}{3} \cdot \frac{y}{x}$ ...(ii)
For tangent of hyperbola:
$\frac{x^2}{4}-\frac{y^2}{3}=1$ ...(iii)
$y^{\prime}=\frac{3}{4} \frac{x}{y}$
Slope of tangent is $m^{\prime}=y^{\prime}=\frac{3}{4} \frac{x}{y}$ ...(iv)
$\therefore$ Normal of ellipse touches the hyperbola. So it becomes tangent to hyperbola.
$\therefore m=m^{\prime} \Rightarrow \frac{4}{3} \frac{y}{x}=\frac{3}{4} \frac{x}{y}$
$\Rightarrow 9 x^2=16 y^2$ ...(v)
Using above value in equation (iii)
$\frac{16 y^2}{9 \times 4}-\frac{y^2}{3}=1 \Rightarrow y^2=9$
From eq ${ }^{\mathrm{n}}$ (iii) $\frac{x^2}{4}-\frac{9}{3}=1 \Rightarrow x^2=16$
Now, square of the slope of the normal is
$m^2=\frac{16}{9} \cdot \frac{y^2}{x^2}=\frac{16}{9} \times \frac{9}{16}=1$
Hence, the correct answer is option: 'A'.