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Ellipse question

2023 · 24 Jan · Shift 1 · Q40
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  5. /2023 · 24 Jan · Shift 1 · Q40

Ellipse question

2023 · 24 Jan · Shift 1 · Q40

JEE MainMathematicsEllipseNumerical+4 / −1
Let C be the largest circle centred at (2, 0) and inscribed in the ellipse x236+y216=1{{{x^2}} \over {36}} + {{{y^2}} \over {16}} = 136x2​+16y2​=1. If (1, α\alphaα) lies on C, then 10 α2\alpha^2α2 is equal to ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 118

  1. Ellipse and circle setup

The ellipse is

x236+y216=1.\frac{x^2}{36}+\frac{y^2}{16}=1.36x2​+16y2​=1.

Its center is (0,0)(0,0)(0,0), with semi-axes a=6a=6a=6 and b=4b=4b=4.

We need the largest circle centered at (2,0)(2,0)(2,0) that lies completely inside the ellipse.

So its radius will be the minimum distance from (2,0)(2,0)(2,0) to the ellipse.

  1. Equation of a point on the ellipse

Let (x,y)(x,y)(x,y) be the point of tangency of the largest such circle with the ellipse. Then the radius is the distance from (2,0)(2,0)(2,0) to (x,y)(x,y)(x,y).

Minimize

D2=(x−2)2+y2D^2=(x-2)^2+y^2D2=(x−2)2+y2

subject to

x236+y216=1.\frac{x^2}{36}+\frac{y^2}{16}=1.36x2​+16y2​=1.

Using the ellipse equation,

y2=16(1−x236)=16−4x29.y^2=16\left(1-\frac{x^2}{36}\right)=16-\frac{4x^2}{9}.y2=16(1−36x2​)=16−94x2​.

So

D2=(x−2)2+16−4x29.D^2=(x-2)^2+16-\frac{4x^2}{9}.D2=(x−2)2+16−94x2​.

Expanding,

D2=x2−4x+4+16−4x29=5x29−4x+20.D^2=x^2-4x+4+16-\frac{4x^2}{9} =\frac{5x^2}{9}-4x+20.D2=x2−4x+4+16−94x2​=95x2​−4x+20.
  1. Minimize the distance

Differentiate with respect to xxx:

ddxD2=10x9−4.\frac{d}{dx}D^2=\frac{10x}{9}-4.dxd​D2=910x​−4.

Set equal to zero:

10x9−4=0  ⟹  10x9=4  ⟹  x=185.\frac{10x}{9}-4=0 \implies \frac{10x}{9}=4 \implies x=\frac{18}{5}.910x​−4=0⟹910x​=4⟹x=518​.

Now find y2y^2y2:

y2=16−49(185)2=16−49⋅32425=16−14425=400−14425=25625.y^2=16-\frac{4}{9}\left(\frac{18}{5}\right)^2 =16-\frac{4}{9}\cdot \frac{324}{25} =16-\frac{144}{25} =\frac{400-144}{25} =\frac{256}{25}.y2=16−94​(518​)2=16−94​⋅25324​=16−25144​=25400−144​=25256​.

So the point of tangency is

(185,±165).\left(\frac{18}{5},\pm \frac{16}{5}\right).(518​,±516​).

Now the minimum distance squared is

r2=(185−2)2+(165)2=(85)2+(165)2=6425+25625=32025=645.r^2=\left(\frac{18}{5}-2\right)^2+\left(\frac{16}{5}\right)^2 =\left(\frac{8}{5}\right)^2+\left(\frac{16}{5}\right)^2 =\frac{64}{25}+\frac{256}{25} =\frac{320}{25} =\frac{64}{5}.r2=(518​−2)2+(516​)2=(58​)2+(516​)2=2564​+25256​=25320​=564​.

Thus the largest circle is

(x−2)2+y2=645.(x-2)^2+y^2=\frac{64}{5}.(x−2)2+y2=564​.
  1. Use the point (1,α)(1,\alpha)(1,α) on the circle

Since (1,α)(1,\alpha)(1,α) lies on the circle,

(1−2)2+α2=645.(1-2)^2+\alpha^2=\frac{64}{5}.(1−2)2+α2=564​.

So

1+α2=645  ⟹  α2=645−1=595.1+\alpha^2=\frac{64}{5} \implies \alpha^2=\frac{64}{5}-1=\frac{59}{5}.1+α2=564​⟹α2=564​−1=559​.

Therefore,

10α2=10⋅595=118.10\alpha^2=10\cdot \frac{59}{5}=118.10α2=10⋅559​=118.
  1. Final answer
118\boxed{118}118​

The derived answer matches the stored correct answer.

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