Let x=4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12 . If P(1,β),β>0 is a point on this ellipse, then the equation of the normal to it at P is:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>4</mn> <mi>x</mi> <mo>-</mo> <mn>2</mn> <mi>y</mi> <mo>=</mo> <mn>1</mn> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>4</mn> <mi>x</mi> <mo>-</mo> <mn>3</mn> <mi>y</mi> <mo>=</mo> <mn>2</mn> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>7</mn> <mi>x</mi> <mo>-</mo> <mn>4</mn> <mi>y</mi> <mo>=</mo> <mn>1</mn> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>8</mn> <mi>x</mi> <mo>-</mo> <mn>2</mn> <mi>y</mi> <mo>=</mo> <mn>5</mn> </math> </span></p>
4 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago
Correct Option - 1
Detailed Solution:

[x]2+2 [x+2]-7=0

[x]2+2 [x]+4-7=0 [x]=1, -3x [1,2) [-3, -2)

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Since  (3,3) lies on x2a2-y2b2=1

9a2-9b2=1

Now, normal at  (3,3) is y-3=-a2b2 (x-3) ,

which passes through  (9,0)b2=2a2

So,  e2=1+b2a2=3

Also,  a2=92

(From (i) and (ii)

Thus,  a2, e2=92, 3

 x2a2+y2b2=1(ab);2b2a=10b2=5a

Now, ?(t)=512+t-t2=812-t-122
?(t)max=812=23=ee2=1-b2a2=49

a2=81 (From (i) and (ii)

So, a2+b2=81+45=126

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve 2025

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