Question

The locus of a point $P(\alpha ,\beta )$ moving under the condition that the line $y=\alpha x+\beta$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

• A. a hyperbola
• B. a parabola
• C. a circle
• D. an ellipse

Answer 1

The correct option is A a hyperbola

Here we are dealing with a locus of moving point $\alpha ,\beta$.

Given that $\alpha ,\beta$ satisfies the condition that $y=\alpha x+\beta$ is a tangent to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

We known the general form of tangent of slope 'm' which is $y=\pm \sqrt{a^2{m}^2-b^2}$

Since $y=\alpha x+\beta$ ia a tangent, we can compare the coefficients of the the 2 equations and equate them if

coefficient of one is the same.Since coefficient of y is 1 in both equations,

$\alpha =m$

$\beta =\sqrt{a^2{m}^2-b^2}$

i.e.,$\beta =\sqrt{a^2{\alpha }^2-b^2}$

i.e.,${\beta }^2=a^2{\alpha }^2-b^2$

$a^2{\alpha }^2-b^2=b^2$

since $\alpha \beta$ is the locus points. this represents the curve,

i.e.,$\frac{x^2}{(\frac{a}{b}{)}^2}-\frac{y^2}{b^2}=1$ which is a hyperbola

Hence option (a) is correct.