Question

Find the points on the ellipse $\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9} = 1$ on which normals are parallel to the line $2x - y = 1$.

Answer 1

Hint: In the given question, we are provided with the equation of an ellipse and a straight line. We have to find the points of the ellipse whose equation is given to us such that the normals are parallel to the given line. So, we first differentiate the equation of the ellipse to find the slope of the normals at any point and then equate it to the slope of the straight line.

Complete step by step solution:
So, the equation of the ellipse is $\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9} = 1$.
Also, the equation of the straight line is $2x - y = 1$.
$ \Rightarrow y = 2x - 1$
So, the slope of the given line is $2$ by comparing the equation with the slope of the intercept form of a straight line.
Now, differentiating the equation of ellipse with respect to x, we get,
$ \Rightarrow \dfrac{{2x}}{4} + \dfrac{{2y}}{9}\dfrac{{dy}}{{dx}} = 0$
Cancelling the common factors in numerator and denominator and shifting the terms of the equation, we get,
$ \Rightarrow \dfrac{{2y}}{9}\dfrac{{dy}}{{dx}} = - \dfrac{x}{2}$
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{9x}}{{4y}}\]
Now, we know that the slope of tangent is $\dfrac{{dy}}{{dx}}$ and slope of normal is $\left( { - \dfrac{{dx}}{{dy}}} \right)$ as the tangent and normal always form a right angle.
So, the slope of normal to the curve is \[\left( { - \dfrac{{dx}}{{dy}}} \right) = \dfrac{{4y}}{{9x}}\] .
Now, equating this with the slope of the line given to us since the normal is parallel to the given line, we get,
 \[ \Rightarrow \dfrac{{4y}}{{9x}} = 2\]
 \[ \Rightarrow y = \dfrac{{9x}}{2}\]
So, we get the value of y as $\dfrac{{9x}}{2}$.
Also, the point lies on the ellipse. So, it would satisfy the equation of the ellipse as well.
Hence, we substitute the value of y as $\dfrac{{9x}}{2}$ in the equation of the ellipse.
$ \Rightarrow \dfrac{{{x^2}}}{4} + \dfrac{{{{\left( {\dfrac{{9x}}{2}} \right)}^2}}}{9} = 1$
Now, evaluating the square and simplifying the expression, we get,
$ \Rightarrow \dfrac{{{x^2}}}{4} + \dfrac{{81{x^2}}}{{4 \times 9}} = 1$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow \dfrac{{{x^2}}}{4} + \dfrac{{9{x^2}}}{4} = 1$
$ \Rightarrow \dfrac{{10}}{4}{x^2} = 1$
Shifting all the constants terms to the right side of the equation and taking square root on both sides if the equation, we get,
$ \Rightarrow x = \pm \sqrt {\dfrac{4}{{10}}} = \pm \dfrac{2}{{\sqrt {10} }}$
Now, substituting the value of x in \[y = \dfrac{{9x}}{2}\] , we get,
 \[ \Rightarrow y = \dfrac{9}{2}\left( { \pm \dfrac{2}{{\sqrt {10} }}} \right)\]
 \[ \Rightarrow y = \pm \dfrac{9}{{\sqrt {10} }}\]
So, the points on the ellipse $\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9} = 1$ on which normals are parallel to the line $2x - y = 1$ are: \[\left( {\dfrac{2}{{\sqrt {10} }},\dfrac{9}{{\sqrt {10} }}} \right)\] and \[\left( { - \dfrac{2}{{\sqrt {10} }}, - \dfrac{9}{{\sqrt {10} }}} \right)\] .

Note: We must know the core concepts and concepts of applications of derivatives in order to solve the given question. We should take care while calculating the slope of normal using differentiation and use the correct formula for the same. We also must know methods of solving equations in order to reach the final answer with the help of simplification rules and transposition.