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Parabola question

2025 · 7 Apr · Shift 1 · Q28
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  5. /2025 · 7 Apr · Shift 1 · Q28

Parabola question

2025 · 7 Apr · Shift 1 · Q28

JEE MainMathematicsParabolaMCQ+4 / −1
Let P be the parabola, whose focus is (−2,1)(-2,1)(−2,1) and directrix is 2x+y+2=02 x+y+2=02x+y+2=0. Then the sum of the ordinates of the points on P, whose abscissa is −-− 2, is
  1. A
    52\frac{5}{2}25​
  2. B
    32\frac{3}{2}23​
  3. C
    34\frac{3}{4}43​
  4. D
    14\frac{1}{4}41​
View written solutionFree

Correct answer: B

  1. Use the definition of parabola

A point Q(x,y)Q(x,y)Q(x,y) lies on the parabola if its distance from the focus equals its perpendicular distance from the directrix.

Given:

  • Focus F(−2,1)F(-2,1)F(−2,1)
  • Directrix 2x+y+2=02x+y+2=02x+y+2=0

So, (x+2)2+(y−1)2=∣2x+y+2∣22+12=∣2x+y+2∣5\sqrt{(x+2)^2+(y-1)^2}=\frac{|2x+y+2|}{\sqrt{2^2+1^2}}=\frac{|2x+y+2|}{\sqrt{5}}(x+2)2+(y−1)2​=22+12​∣2x+y+2∣​=5​∣2x+y+2∣​

Squaring both sides, (x+2)2+(y−1)2=(2x+y+2)25(x+2)^2+(y-1)^2=\frac{(2x+y+2)^2}{5}(x+2)2+(y−1)2=5(2x+y+2)2​

  1. Substitute the given abscissa x=−2x=-2x=−2

We need the points on the parabola whose abscissa is −2-2−2, so put x=−2x=-2x=−2.

Then (−2+2)2+(y−1)2=(2(−2)+y+2)25( -2+2)^2+(y-1)^2=\frac{(2(-2)+y+2)^2}{5}(−2+2)2+(y−1)2=5(2(−2)+y+2)2​ 0+(y−1)2=(y−2)250+(y-1)^2=\frac{(y-2)^2}{5}0+(y−1)2=5(y−2)2​

Thus, 5(y−1)2=(y−2)25(y-1)^2=(y-2)^25(y−1)2=(y−2)2

  1. Expand and simplify

5(y2−2y+1)=y2−4y+45(y^2-2y+1)=y^2-4y+45(y2−2y+1)=y2−4y+4 5y2−10y+5=y2−4y+45y^2-10y+5=y^2-4y+45y2−10y+5=y2−4y+4 4y2−6y+1=04y^2-6y+1=04y2−6y+1=0

  1. Find the sum of ordinates

The two ordinates are the roots of 4y2−6y+1=04y^2-6y+1=04y2−6y+1=0

For a quadratic ay2+by+c=0ay^2+by+c=0ay2+by+c=0, sum of roots is −ba-\frac{b}{a}−ab​.

Hence, y1+y2=−−64=64=32y_1+y_2=-\frac{-6}{4}=\frac{6}{4}=\frac{3}{2}y1​+y2​=−4−6​=46​=23​

  1. Compare with options

Therefore, the required sum is 32\boxed{\frac{3}{2}}23​​

So the correct option is B.

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