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Parabola question

2007 · Shift 1 · Q41
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Parabola question

2007 · Shift 1 · Q41

JEE AdvancedMathematicsParabolaMCQ+3 / −1
Consider the circle x2+y2=9{x^2} + {y^2} = 9x2+y2=9 and the parabola y2=8x{y^2} = 8xy2=8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S.The radius of the incircle of the triangle PQR is
  1. A
    4 units
  2. B
    3 units
  3. C
    83\frac{8}{3}38​ units
  4. D
    2 units
View written solutionFree

Correct answer: D

  1. Find the intersection points PPP and QQQ

We have x2+y2=9x^2+y^2=9x2+y2=9 and y2=8x.y^2=8x.y2=8x.

Substitute x=y28x=\dfrac{y^2}{8}x=8y2​ into the circle: (y28)2+y2=9.\left(\frac{y^2}{8}\right)^2+y^2=9.(8y2​)2+y2=9.

So, y464+y2=9\frac{y^4}{64}+y^2=964y4​+y2=9 y4+64y2−576=0.y^4+64y^2-576=0.y4+64y2−576=0.

Let t=y2t=y^2t=y2. Then t2+64t−576=0.t^2+64t-576=0.t2+64t−576=0.

Solving, t=8(positive root).t=8 \quad (\text{positive root}).t=8(positive root).

Hence, y2=8  ⟹  y=±22.y^2=8 \implies y=\pm 2\sqrt2.y2=8⟹y=±22​. Then x=y28=1.x=\frac{y^2}{8}=1.x=8y2​=1.

Therefore, P=(1,22),Q=(1,−22).P=(1,2\sqrt2), \qquad Q=(1,-2\sqrt2).P=(1,22​),Q=(1,−22​).

  1. Find point RRR where tangents to the circle at PPP and QQQ meet the xxx-axis

For the circle x2+y2=9x^2+y^2=9x2+y2=9, tangent at (x1,y1)(x_1,y_1)(x1​,y1​) is xx1+yy1=9.xx_1+yy_1=9.xx1​+yy1​=9.

At P=(1,22)P=(1,2\sqrt2)P=(1,22​): x+22 y=9.x+2\sqrt2\,y=9.x+22​y=9.

Its intersection with the xxx-axis is obtained by setting y=0y=0y=0: x=9.x=9.x=9. So tangent at PPP meets the xxx-axis at (9,0)(9,0)(9,0).

At Q=(1,−22)Q=(1,-2\sqrt2)Q=(1,−22​): x−22 y=9,x-2\sqrt2\,y=9,x−22​y=9, which also meets the xxx-axis at (9,0).(9,0).(9,0).

Thus, R=(9,0).R=(9,0).R=(9,0).

  1. Find point SSS where tangents to the parabola at PPP and QQQ meet the xxx-axis

For parabola y2=8xy^2=8xy2=8x, we have 4a=8  ⟹  a=24a=8 \implies a=24a=8⟹a=2.

Tangent to y2=4axy^2=4axy2=4ax at (x1,y1)(x_1,y_1)(x1​,y1​) is yy1=2a(x+x1).yy_1=2a(x+x_1).yy1​=2a(x+x1​). Since a=2a=2a=2, yy1=4(x+x1).yy_1=4(x+x_1).yy1​=4(x+x1​).

At P=(1,22)P=(1,2\sqrt2)P=(1,22​): 22 y=4(x+1).2\sqrt2\,y=4(x+1).22​y=4(x+1). Set y=0y=0y=0 to get its xxx-intercept: 0=4(x+1)  ⟹  x=−1.0=4(x+1) \implies x=-1.0=4(x+1)⟹x=−1.

At Q=(1,−22)Q=(1,-2\sqrt2)Q=(1,−22​): −22 y=4(x+1),-2\sqrt2\,y=4(x+1),−22​y=4(x+1), and again with y=0y=0y=0, x=−1.x=-1.x=−1.

So, S=(−1,0).S=(-1,0).S=(−1,0).

  1. Understand triangle PQRPQRPQR

The vertices are P=(1,22),Q=(1,−22),R=(9,0).P=(1,2\sqrt2),\quad Q=(1,-2\sqrt2),\quad R=(9,0).P=(1,22​),Q=(1,−22​),R=(9,0).

Now compute side lengths:

  • PQPQPQ is vertical: PQ=22−(−22)=42.PQ=2\sqrt2-(-2\sqrt2)=4\sqrt2.PQ=22​−(−22​)=42​.

  • PRPRPR:

=\sqrt{8^2+(2\sqrt2)^2} =\sqrt{64+8} =\sqrt{72}=6\sqrt2.$$ - Similarly, $$QR=6\sqrt2.$$ So triangle $PQR$ is isosceles with sides $$6\sqrt2,\ 6\sqrt2,\ 4\sqrt2.$$ --- 5. **Compute area of triangle $PQR$** Take base $PQ=4\sqrt2$. Since $PQ$ is the vertical line $x=1$, the perpendicular distance of $R=(9,0)$ from this line is $$9-1=8.$$ Therefore, $$\Delta = \frac12 \times 4\sqrt2 \times 8 = 16\sqrt2.$$ --- 6. **Compute semiperimeter** $$s=\frac{PQ+PR+QR}{2} =\frac{4\sqrt2+6\sqrt2+6\sqrt2}{2} =\frac{16\sqrt2}{2}=8\sqrt2.$$ --- 7. **Find the inradius** Using $$\Delta = rs,$$ we get $$r=\frac{\Delta}{s} = \frac{16\sqrt2}{8\sqrt2}=2.$$ Thus, the radius of the incircle of triangle $PQR$ is $$\boxed{2}.$$ So the correct option is **D**. --- 8. **Comparison with stored answer** Stored correct answer: **D** Derived answer: **D** They agree.
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