Question

plz solve Q no.-8 from Conic section ellipse

Answer 1

Step 1 :

It is given that the centre is at (0,0) and the major axis is on the y - axis.

Hence the equation of the ellipse will be of the form x2b2x2b2+y2a2+y2a2=1=1

The ellipse passes through the points (3, 2) and (1, 6)

Substituting these values for x and y we get,

Step 2 :

9b29b2+4a2+4a2=1=1

⇒9a2+4b2=a2b2⇒9a2+4b2=a2b2--------(1)

and 1b21b2+36a2+36a2=1=1

a2+36b2=a2b2a2+36b2=a2b2------(2)

Let us solve equation (1) and (2) to obtain the values for a2a2 and b2b2

9a2+4b2=a2b2(×9)9a2+4b2=a2b2(×9)

a2+36b2=a2b2a2+36b2=a2b2

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81a2+36b2=9a2b281a2+36b2=9a2b2

a2+36b2=a2b2a2+36b2=a2b2

(−)(−)(−)(−)(−)(−)

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80a2=8a2b280a2=8a2b2

∴b2=10∴b2=10

and a2=40a2=40

Hence the equation of the ellipse is

x210x210+y240+y240=1=1

or 4x2+y2=40