Question

Find the equation of the ellipse in standard form if passing through the points $(- 3, 1)$ and $(2, - 2)$ .

Accepted Answer

Let the required equation of the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , where $a > b$

Since the ellipse passes through the points $(- 3, 1)$ and $(2, - 2)$ . So, the co-ordinates of the point satisfies its equation.

Substituting $x = - 3$ and $y = 1$ , we get

$\frac{{(- 3)}^{2}}{a^{2}} + \frac{1}{b^{2}} = 1$

$\Rightarrow \frac{9}{a^{2}} + \frac{1}{b^{2}} = 1$ ...... $(1)$

Now, substituting $x = 2$ and $y = - 2$ in the standard equation of the ellipse, we get

$\frac{2^{2}}{a^{2}} + \frac{{(- 2)}^{2}}{b^{2}} = 1$

$\Rightarrow \frac{4}{a^{2}} + \frac{4}{b^{2}} = 1$ ........... $(2)$

Put $\frac{1}{a^{2}} = A$ and $\frac{1}{b^{2}} = B$

Thus, from equation $(1)$ , we have

$9 A + B = 1$ ..... $(3)$

And, from equation $(2)$ , we have

$4 A + 4 B = 1$ ...... $(4)$

Multiplying equation $(3)$ by $4$ and subtracting equation $(4)$ , we get

$32 A = 3$

$\Rightarrow A = \frac{3}{32}$

Substituting $A = \frac{3}{32}$ in equation $(4)$ , we get

$4 \times \frac{3}{32} + 4 B = 1$

$\Rightarrow B = \frac{5}{32}$

Thus, $A = \frac{3}{32} = \frac{1}{a^{2}}$

$\Rightarrow a^{2} = \frac{32}{3}$ and

$B = \frac{5}{32} = \frac{1}{b^{2}}$

$\Rightarrow b^{2} = \frac{32}{5}$

Hence, the required equation of the ellipse is $\frac{x^{2}}{\frac{32}{3}} + \frac{y^{2}}{\frac{32}{5}} = 1$

$\Rightarrow 3 x^{2} + 5 y^{2} = 32$

Maharashtra Board Solutions for Chapter: Conic Sections, Exercise 2: EXERCISE 7.2

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Conic Sections, Exercise 2: EXERCISE 7.2

Questions from Maharashtra Board Solutions for Chapter: Conic Sections, Exercise 2: EXERCISE 7.2 with Hints & Solutions

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