Question

What is the equation of the normal to the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ at the point $(5\sqrt{3},2\sqrt{2})$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Here we can either the formula or find it using the slope-point method for getting a line's equation.

Normal is usually found by the methodology in which we find the slope of the line by using derivatives. We know that tangent and normal at a point will be perpendicular. Differentiating the equation of hyperbola with respect to x.

$\frac{2x}{25}-\frac{2y}{16}{y}^{\prime }=1$

At the point ($5\sqrt{3},2\sqrt{2})$

$\frac{2.5\sqrt{3}}{25}-\frac{2.2\sqrt{2}}{16}{y}^{\prime }=1$

$y^{\prime }=[1-\frac{2\sqrt{3}}{5}].2(-1)\sqrt{2}$

$=\frac{(2\sqrt{3}-5)2\sqrt{2}}{5}$

Slope of normal is,

$m=-\frac{1}{y^{\prime }}=\frac{5}{2\sqrt{2}.(5-2\sqrt{3})}$

$\therefore$ equation of the normal is,

$y-2\sqrt{2}=\frac{5}{2\sqrt{2}.(5-2\sqrt{3})}.(x-5\sqrt{3})$

$\left(2\sqrt{2}\right)[5-2\sqrt{3}]y-8.(5-2\sqrt{3})=5x-25\sqrt{3}$

$i.e.,5x+y.2\sqrt{2}(2\sqrt{3}-5)+40-16\sqrt{3}-25\sqrt{3}=0$

$i.e.,5x+y.2\sqrt{2}(2\sqrt{3}-5)+40-41\sqrt{3}=0$

$i.e.,5\sqrt{2}x+8\sqrt{3}y-41\sqrt{6}=0$

If we derive for a general case, i.e., for a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{1b^2}=1$

at a point $(x_1,y_1)$ we will get the formula for normal as,

$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

Here, $\frac{{25}^{5}x}{5\sqrt{3}}+\frac{{16}^8.y}{2\sqrt{2}}=41$

$5\sqrt{2}x+8\sqrt{3}y-41\sqrt{6}=0$