Question

The distance of the point \[^{\prime}\theta^{\prime}\] on the ellipse \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\] from a focus is______.

विकल्प

• \[\text{a}(\text{e}+\cos\theta)\]

• \[\text{a}(\text{e}-\cos\theta)\]

• \[a(1+e\cos\theta)\]

• \[\text{a}(1+2\text{e}\cos\theta)\]

Answer 1

The distance of the point \[^{\prime}\theta^{\prime}\] on the ellipse \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\] from a focus is \[a(1+e\cos\theta)\].

Explanation:

For the ellipse, the parametric point is \[P=(a\cos\theta,b\sin\theta).\]

The focal distance formula from focus\[(ae,0)\] is:

\[SP=a-ex_1=a-e(a\cos\theta)=a(1-e\cos\theta)\]

And from the other focus \[(-ae,0)\]:

\[S^{\prime}P=a+e(a\cos\theta)=a(1+e\cos\theta)\]