Question

What is the eccentric angle in the first quadrant of a point on the ellipse ${\displaystyle \frac{x^2}{10}}+{\displaystyle \frac{y^2}{8}}=1$ at a distance 3 units from the centre of the ellipse?
(A). ${\displaystyle \frac{\pi }{6}}$
(B). ${\displaystyle \frac{\pi }{4}}$
(C). ${\displaystyle \frac{\pi }{3}}$
(D). ${\displaystyle \frac{\pi }{2}}$

Answer 1

Hint- In such types of questions, just follow the simple approach first take the parametric equation of ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ centred at origin and find the distance of $x=a\cos \theta ,y=b\sin \theta$ point in the first quadrant from the origin and then equate that to the given distance to get the required value of $\theta$ i.e. the eccentric angle.

Complete step-by-step solution -
Let us suppose the point $$P$$ is on the ellipse given.

Now we have $a^2=10,b^2=8$ so we have $a=\sqrt{10},b=\sqrt{8}$
We have the parametric equation of the ellipse as $x=a\cos \theta ,y=b\sin \theta$ and here $x=\sqrt{10}\cos \theta ,y=\sqrt{8}\sin \theta$
Now, we know that the distance formula to calculate distance between two points say A $(x_1,y_1)$ and B $(x_2,y_2)$ which is $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Here the two points are $P(\sqrt{10}\cos \theta ,\sqrt{8}\sin \theta )$ and O (0, 0)
So, $d=\sqrt{{\left(0-\sqrt{10}\cos \theta \right)}^2+{\left(0-\sqrt{8}\sin \theta \right)}^2}$
$\Rightarrow d=\sqrt{{\left(-\sqrt{10}\cos \theta \right)}^2+{\left(-\sqrt{8}\sin \theta \right)}^2}$
$$\Rightarrow d=\sqrt{10{\cos }^2\theta +8{\sin }^2\theta }$$
Now, d = 3 as given in the question so we get,
$$\Rightarrow 3=\sqrt{10{\cos }^2\theta +8{\sin }^2\theta }$$
Squaring both sides, we get,
$$\Rightarrow 3^2=10{\cos }^2\theta +8{\sin }^2\theta$$
$$\Rightarrow 9=10{\cos }^2\theta +8{\sin }^2\theta$$
Now we know that ${\cos }^2\theta =1-{\sin }^2\theta$
$$\Rightarrow 9=10(1-{\sin }^2\theta )+8{\sin }^2\theta$$
$$\Rightarrow 9=10-10{\sin }^2\theta +8{\sin }^2\theta$$
$$\Rightarrow 10{\sin }^2\theta -8{\sin }^2\theta =10-9$$
$$\Rightarrow 2{\sin }^2\theta =1$$
$$\Rightarrow {\sin }^2\theta ={\displaystyle \frac{1}{2}}$$
$$\Rightarrow \sin \theta ={\displaystyle \frac{1}{\sqrt{2}}}$$
We know that $\sin \left({\displaystyle \frac{\pi }{4}}\right)={\displaystyle \frac{1}{\sqrt{2}}}$ in the first quadrant.
So, we have $\theta ={\displaystyle \frac{\pi }{4}}$
Hence, the eccentric angle in the first quadrant of a point on the ellipse ${\displaystyle \frac{x^2}{10}}+{\displaystyle \frac{y^2}{8}}=1$ at a distance 3 units from the centre of the ellipse is $\theta ={\displaystyle \frac{\pi }{4}}$
$\therefore$ Option B. ${\displaystyle \frac{\pi }{4}}$ is the correct answer.

Note- In such types of questions, just keep in mind the distance formula to calculate the distance between two points i.e. for two points say two points say A $(x_1,y_1)$ and B $(x2,y2)$ distance is $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ and also keep in mind the parametric equation of ellipse as $x=a\cos \theta ,y=b\sin \theta$.