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Ellipse question

2021 · 27 Jul · Shift 1 · Q28
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  5. /2021 · 27 Jul · Shift 1 · Q28

Ellipse question

2021 · 27 Jul · Shift 1 · Q28

JEE MainMathematicsEllipseMCQ+4 / −1
A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity 13{1 \over 3}31​ and the distance of the nearer focus from this directrix is 853{8 \over {\sqrt {53} }}53​8​, then the equation of the other directrix can be :
  1. A
    11x + 7y + 8 = 0 or 11x + 7y −-− 15 = 0
  2. B
    11x −-− 7y −-− 8 = 0 or 11x + 7y + 15 = 0
  3. C
    2x −-− 7y + 29 = 0 or 2x −-− 7y −-− 7 = 0
  4. D
    2x −-− 7y −-− 39 = 0 or 2x −-− 7y −-− 7 = 0
View written solutionFree

Correct answer: C

  1. Find the reflected ray

The ray comes from (2,1)(2,1)(2,1), reflects at a point PPP on the yyy-axis, and then passes through (5,3)(5,3)(5,3).

Since reflection occurs on the yyy-axis, the incident ray from (2,1)(2,1)(2,1) can be replaced by a straight line from the reflection of (2,1)(2,1)(2,1) in the yyy-axis, i.e. from (−2,1)(-2,1)(−2,1) to (5,3)(5,3)(5,3).

So the reflected ray lies on the same line as the line joining (−2,1)(-2,1)(−2,1) and (5,3)(5,3)(5,3).

Slope of this line is

m=3−15−(−2)=27.m=\frac{3-1}{5-(-2)}=\frac{2}{7}.m=5−(−2)3−1​=72​.

Hence equation is

y−1=27(x+2)y-1=\frac{2}{7}(x+2)y−1=72​(x+2) 7y−7=2x+47y-7=2x+47y−7=2x+4 2x−7y+11=0.2x-7y+11=0.2x−7y+11=0.

Thus one directrix of the ellipse is

2x−7y+11=0.2x-7y+11=0.2x−7y+11=0.
  1. Use eccentricity and focus-directrix distance

Given eccentricity

e=13.e=\frac13.e=31​.

For an ellipse, if the distance between the two directrices is 2ae\dfrac{2a}{e}e2a​ and the distance between the foci is 2ae2ae2ae, then the perpendicular distance from a focus to the nearer directrix is

ae−ae=a(1e−e).\frac{a}{e}-ae = a\left(\frac1e-e\right).ea​−ae=a(e1​−e).

Substitute e=13e=\frac13e=31​:

a(3−13)=a⋅83.a\left(3-\frac13\right)=a\cdot \frac{8}{3}.a(3−31​)=a⋅38​.

This is given as

853.\frac{8}{\sqrt{53}}.53​8​.

So

a⋅83=853a\cdot \frac{8}{3}=\frac{8}{\sqrt{53}}a⋅38​=53​8​ a=353.a=\frac{3}{\sqrt{53}}.a=53​3​.

Therefore distance between the two directrices is

2ae=2⋅(3/53)1/3=1853.\frac{2a}{e}=\frac{2\cdot (3/\sqrt{53})}{1/3}=\frac{18}{\sqrt{53}}.e2a​=1/32⋅(3/53​)​=53​18​.
  1. Find the other directrix

The given directrix is

2x−7y+11=0.2x-7y+11=0.2x−7y+11=0.

Any line parallel to it has form

2x−7y+k=0.2x-7y+k=0.2x−7y+k=0.

Distance between

2x−7y+11=02x-7y+11=02x−7y+11=0

and

2x−7y+k=02x-7y+k=02x−7y+k=0

is

∣k−11∣22+(−7)2=∣k−11∣53.\frac{|k-11|}{\sqrt{2^2+(-7)^2}}=\frac{|k-11|}{\sqrt{53}}.22+(−7)2​∣k−11∣​=53​∣k−11∣​.

Set this equal to 1853\dfrac{18}{\sqrt{53}}53​18​:

∣k−11∣53=1853\frac{|k-11|}{\sqrt{53}}=\frac{18}{\sqrt{53}}53​∣k−11∣​=53​18​ ∣k−11∣=18.|k-11|=18.∣k−11∣=18.

Thus

k=29ork=−7.k=29 \quad \text{or} \quad k=-7.k=29ork=−7.

Hence the other directrix can be

2x−7y+29=02x-7y+29=02x−7y+29=0

or

2x−7y−7=0.2x-7y-7=0.2x−7y−7=0.
  1. Match with options

This is exactly Option C.

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.

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