Question

If the line $x+y=0$ touches the curve $2y^2=\alpha x^2+\beta$ at $(1,-1),$ then $(\alpha ,\beta )=$

• A. $(-2,4)$
• B. $(-1,3)$
• C. $(4,-2)$
• D. $(2,0)$

Answer 1

The correct option is D $(2,0)$

Given equation of curve is,

$2y^2=\alpha x^2+\beta$

Point $(1,-1)$ lies on the curve. Thus, it must satisfy given equation of curve.

$\therefore 2{(-1)}^2=\alpha {\left(1\right)}^2+\beta$

$\therefore 2\left(1\right)=\alpha \left(1\right)+\beta$

$\therefore 2=\alpha +\beta$ (1)

Now, equation of tangent to curve is,

$x+y=0$

$\therefore y=-x$

Thus, slope of tangent is, $m_1=-1$ (2)

Equation of the curve is,

$2y^2=\alpha x^2+\beta$

Differentiate w.r.t. x, we get,

$2\times 2y\frac{dy}{dx}=2\alpha x+0$

$\therefore 4y\frac{dy}{dx}=2\alpha x$

$\therefore \frac{dy}{dx}=\frac{2\alpha x}{4y}$

Slope of curve at $(1,-1)$ is,

$m_2=\frac{2\alpha \times 1}{4\times -1}$

$m_2=\frac{2\alpha }{-4}$

$\therefore m_2=\frac{\alpha }{-2}$ (3)

At point $(1,-1)$, slope of tangent and normal will be same.

Thus, from equation (2) and (3),

$\frac{\alpha }{-2}=-1$

$\therefore \alpha =2$

Put this value in equation (1), we get,

$2+\beta =2$

$\therefore \beta =0$

$\therefore (\alpha ,\beta )=(2,0)$