Question

The equation of the tangent at $P\left( \theta \right)$ i.e., $P\left( {a\cos \theta ,b\sin \theta } \right)$ to the ellipse, $S = 0$ is $\dfrac{{x\cos \theta }}{a} + \dfrac{{y\sin \theta }}{b} = 1$.

Answer 1

Hint: First, find the derivative of the equation of the ellipse. Then, find the slope of the tangent at the given point. After that find the equation of the tangent at the given point. After further calculation, the desired output will come out.

Complete step-by-step answer:
Given: - The point $P\left( {a\cos \theta ,b\sin \theta } \right)$.
The standard equation of the ellipse is,
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Now differentiate the above equation on both sides with respect to x, we get,
$\dfrac{{2x}}{{{a^2}}} + \dfrac{{2y}}{{{b^2}}}\dfrac{{dy}}{{dx}} = 0$
Move the x part to the right side of the equation,
$\dfrac{{2y}}{{{b^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{{a^2}}}$
Multiply both sides by $\dfrac{{{b^2}}}{{2y}}$ to make the coefficient of $\dfrac{{dy}}{{dx}}$ equal to 1,
$\dfrac{{{b^2}}}{{2y}} \times \dfrac{{2y}}{{{b^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{{a^2}}} \times \dfrac{{{b^2}}}{{2y}}$
Cancel out the common factor from the numerator and denominator on the left side and multiply on the right side,
$\dfrac{{dy}}{{dx}} = - \dfrac{{{b^2}x}}{{{a^2}y}}$ ….. (1)
Let m be the slope of the tangent at the given point $P\left( {a\cos \theta ,b\sin \theta } \right)$. Then,
$m = \dfrac{{dy}}{{dx}}$
From equation (1),
$m = - \dfrac{{{b^2}x}}{{{a^2}y}}$
Put $x = a\cos \theta $ and $y = b\sin \theta $,
$m = - \dfrac{{{b^2} \times a\cos \theta }}{{{a^2} \times b\sin \theta }}$
Cancel out the common factors from numerator and denominator,
$m = - \dfrac{{b\cos \theta }}{{a\sin \theta }}$
The equation of the tangent at the given point $P\left( {a\cos \theta ,b\sin \theta } \right)$. Then,
$y - b\sin \theta = - \dfrac{{b\cos \theta }}{{a\sin \theta }}\left( {x - a\cos \theta } \right)$
Cross multiply the term,
$\left( {y - b\sin \theta } \right)a\sin \theta = a\cos \theta \left( {x - a\cos \theta } \right)$
Divide both sides by ab and open brackets and multiply the terms,
$\dfrac{{ay\sin \theta }}{{ab}} - \dfrac{{ab{{\sin }^2}\theta }}{{ab}} = - \dfrac{{bx\cos \theta }}{{ab}} + \dfrac{{ab{{\cos }^2}\theta }}{{ab}}$
Cancel out the common factors from numerator and denominator on both sides of the equation,
$\dfrac{{y\sin \theta }}{b} - {\sin ^2}\theta = - \dfrac{{x\cos \theta }}{a} + {\cos ^2}\theta $
Move the constant term on the right side and variable term on the left side,
$\dfrac{{x\cos \theta }}{a} + \dfrac{{y\sin \theta }}{b} = {\sin ^2}\theta + {\cos ^2}\theta $
Since, ${\sin ^2}\theta + {\cos ^2}\theta = 1$, Then,
$\dfrac{{x\cos \theta }}{a} + \dfrac{{y\sin \theta }}{b} = 1$

Hence, the equation of the tangent at $P\left( \theta \right)$to the ellipse is $\dfrac{{x\cos \theta }}{a} + \dfrac{{y\sin \theta }}{b} = 1$.

Note: The students might make mistakes by not differentiating the equation with respect to x.
$\dfrac{{2x}}{{{a^2}}} + \dfrac{{2y}}{{{b^2}}}\dfrac{{dy}}{{dx}} = 0$
An ellipse is a shape that looks like an oval or a flattened circle.
In geometry, an ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve.
The equation of the ellipse is: $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}}$
where, $\left( {h,k} \right)$ is the center of the ellipse.