Question

Find the locus of point of intersection of perpendicular tangents to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$

• A. x² + y² = 16
• B. x² + y² = 7
• C. x² + y² = 23
• D. x² + y² = 9

Answer 1

The correct option is B

x² + y² = 7

Let p(h,k) be the point of intersection of two perpendicular tangents.

Equation of pair of tangents is $S{S}_1=T^2$

$S=\frac{x^2}{16}-\frac{y^2}{9}-1$

$S_1=\frac{h^2}{16}-\frac{y^2}{9}-1$

$T=\frac{hx}{16}-\frac{ky}{9}-1$

$S{S}_1=T^2$

$\Rightarrow (\frac{x^2}{16}-\frac{y^2}{9}-1)(\frac{h^2}{16}-\frac{k^2}{9}-1)={(\frac{hx}{16}-\frac{ky}{9}-1)}^2$

$\Rightarrow \frac{x^2}{16}(\frac{h^2}{16}-\frac{k^2}{9}-1)-\frac{y^2}{9}(\frac{h^2}{16}-\frac{k^2}{9}-1)-(\frac{h^2}{16}-\frac{k^2}{9}-1)$

$=\frac{h^2{x}^2}{{16}^2}+\frac{k^2{y}^2}{{16}^2}+1-\frac{2hkxy}{16\times 9}+\frac{2ky}{9}+\frac{2hx}{16}$

$\frac{x^2}{16}(-\frac{k^2}{9}-1)-\frac{y^2}{9}(\frac{h^2}{16}-1)+\frac{hkxy}{72}-\frac{2kg}{9}-\frac{hx}{8}-(\frac{h^2}{16}-\frac{k^2}{9})=0$ ......(i)

Since, equation (i), represents two perpendicular lines

coefficient of $x^2$+coefficient of $y^2$=0

$\frac{1}{16}(-\frac{k^2}{9}-1)-\frac{1}{9}(\frac{h^2}{16}-1)=0$

$-k^2-9-(h^2-16)=0$

$-k^2-9-h^2+16=0$

$k^2+h^2-7=0$

$k^2+h^2=7$

For required locus replacing $h=x\&k=y$

$\Rightarrow y^2+x^2=7or{x}^2+y^2=7$