Question

The locus of the point of intersection of tangents to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ which meet at a constant angle $\beta$, will be

• A. $x^2+y^2+a^2+b^2$ if$\beta ={90}^o$
• B. $(x^2+y^2+b^2-a^2{)}^2=4(a^2{y}^2-b^2{x}^2+a^2{b}^2)$ if$\beta ={45}^o$
• C. If the locus passes through (0, a) ; $ta{n}^2\beta =\frac{4a^2(a^2+b^2)}{b^4}$
• D. If the locus passes through origin $co{t}^2\beta =\frac{(b^2-a^2{)}^2}{4a^2{b}^2}$

Answer 1

Answer: (B), (C), (D)

If the locus passes through (0, a) ; $ta{n}^2\beta =\frac{4a^2(a^2+b^2)}{b^4}$
$(x^2+y^2+b^2-a^2{)}^2=4(a^2{y}^2-b^2{x}^2+a^2{b}^2)$ if$\beta ={45}^o$
If the locus passes through origin $co{t}^2\beta =\frac{(b^2-a^2{)}^2}{4a^2{b}^2}$

Let the point of intersection be (h, k)

so $S{S}_1=T^2$

$[\frac{x^2}{a^2}-\frac{y^2}{b^2}-1][\frac{h^2}{a^2}-\frac{k^2}{b^2}-1]={[\frac{xh}{a^2}-\frac{yk}{b^2}-1]}^2$

so, $tan\beta =\left|\frac{2\sqrt{h^2-AB}}{A+B}\right|$

$\Rightarrow tan\beta =\frac{2\sqrt{\frac{h^2{k}^2}{a^4{b}^4}+\left[\frac{k^2+h^2}{a^2{b}^2}\right]\left[\frac{a^2-h^2}{a^2{b}^2}\right]}}{|-\frac{h^2}{a^2{h}^2}+\frac{1}{h^2}-\frac{k^2}{a^2{k}^2}-\frac{1}{a^2}|}$

$\Rightarrow tan\beta =\frac{2\sqrt{a^2{y}^2-b^2{x}^2+a^2{b}^2}}{|a^2-b^2-x^2-y^2|}$

$\Rightarrow (a^2{y}^2+a^2{b}^2-b^2{x}^2)=\frac{ta{n}^2\beta }{4}(x^2+y^2+b^2-a^2{)}^2$