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Parabola question

2025 · 8 Apr · Shift 2 · Q48
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Parabola question

2025 · 8 Apr · Shift 2 · Q48

JEE MainMathematicsParabolaNumerical+4 / −1
Let rrr be the radius of the circle, which touches xxx- axis at point (a,0),a<0(a, 0), a\lt 0(a,0),a<0 and the parabola y2=9x\mathrm{y}^2=9 xy2=9x at the point (4,6)(4,6)(4,6). Then rrr is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 30

  1. Set up the circle

Since the circle touches the xxx-axis at (a,0)(a,0)(a,0), its center must lie vertically above or below this point by a distance equal to the radius.

Because the point of tangency with the parabola is (4,6)(4,6)(4,6) (which is above the xxx-axis), the relevant circle has center

C=(a,r)C=(a,r)C=(a,r)

and radius rrr.

So the circle passes through (4,6)(4,6)(4,6) and is tangent to the parabola y2=9xy^2=9xy2=9x at that point.

  1. Use the tangent condition with the parabola

For the parabola

y2=9x,y^2=9x,y2=9x,

we compare with the standard form y2=4axy^2=4axy2=4ax. Hence,

4a=9⇒a=94.4a=9 \quad \Rightarrow \quad a=\frac94.4a=9⇒a=49​.

For y2=4axy^2=4axy2=4ax, the tangent at parameter ttt is at point (at2,2at)(at^2,2at)(at2,2at).

Given point (4,6)(4,6)(4,6):

2at=6⇒2⋅94⋅t=62at=6 \Rightarrow 2\cdot \frac94 \cdot t=62at=6⇒2⋅49​⋅t=6 92t=6⇒t=43.\frac92 t=6 \Rightarrow t=\frac{4}{3}.29​t=6⇒t=34​.

This is consistent since

at2=94⋅169=4.at^2=\frac94\cdot \frac{16}{9}=4.at2=49​⋅916​=4.

The tangent to y2=4axy^2=4axy2=4ax at parameter ttt has slope

1t.\frac{1}{t}.t1​.

So slope of tangent at (4,6)(4,6)(4,6) is

14/3=34.\frac{1}{4/3}=\frac34.4/31​=43​.

Therefore the normal slope is

−43.-\frac{4}{3}.−34​.

Since the circle is tangent to the parabola at (4,6)(4,6)(4,6), the radius to the point of contact lies along the normal. Hence the center (a,r)(a,r)(a,r) lies on the normal through (4,6)(4,6)(4,6).

Equation of normal:

y−6=−43(x−4).y-6=-\frac43(x-4).y−6=−34​(x−4).

Substitute (x,y)=(a,r)(x,y)=(a,r)(x,y)=(a,r):

r−6=−43(a−4).r-6=-\frac43(a-4).r−6=−34​(a−4).

So,

3r−18=−4a+163r-18=-4a+163r−18=−4a+16 4a+3r=34.(1)4a+3r=34. \qquad (1)4a+3r=34.(1)
  1. Use the fact that (4,6)(4,6)(4,6) lies on the circle

Distance from center (a,r)(a,r)(a,r) to (4,6)(4,6)(4,6) equals radius rrr:

(4−a)2+(6−r)2=r2.(4-a)^2+(6-r)^2=r^2.(4−a)2+(6−r)2=r2.

Expanding,

(a−4)2+r2−12r+36=r2(a-4)^2 + r^2-12r+36 = r^2(a−4)2+r2−12r+36=r2 (a−4)2−12r+36=0(a-4)^2-12r+36=0(a−4)2−12r+36=0 (a−4)2=12r−36.(2)(a-4)^2=12r-36. \qquad (2)(a−4)2=12r−36.(2)
  1. Solve using (1)

From (1),

4a=34−3r⇒a=34−3r4.4a=34-3r \Rightarrow a=\frac{34-3r}{4}.4a=34−3r⇒a=434−3r​.

Then

a−4=34−3r4−4=18−3r4=3(6−r)4.a-4=\frac{34-3r}{4}-4=\frac{18-3r}{4}=\frac{3(6-r)}{4}.a−4=434−3r​−4=418−3r​=43(6−r)​.

So,

(a−4)2=9(6−r)216.(a-4)^2=\frac{9(6-r)^2}{16}.(a−4)2=169(6−r)2​.

Using (2):

9(6−r)216=12r−36.\frac{9(6-r)^2}{16}=12r-36.169(6−r)2​=12r−36.

Multiply by 161616:

9(6−r)2=192r−576.9(6-r)^2=192r-576.9(6−r)2=192r−576.

Expand:

9(r2−12r+36)=192r−5769(r^2-12r+36)=192r-5769(r2−12r+36)=192r−576 9r2−108r+324=192r−5769r^2-108r+324=192r-5769r2−108r+324=192r−576 9r2−300r+900=0.9r^2-300r+900=0.9r2−300r+900=0.

Divide by 333:

3r2−100r+300=0.3r^2-100r+300=0.3r2−100r+300=0.

Solve:

r=100±10000−36006=100±806.r=\frac{100\pm \sqrt{10000-3600}}{6} =\frac{100\pm 80}{6}.r=6100±10000−3600​​=6100±80​.

Thus,

r=30orr=103.r=30 \quad \text{or} \quad r=\frac{10}{3}.r=30orr=310​.
  1. Choose the correct radius using a<0a<0a<0

From

a=34−3r4,a=\frac{34-3r}{4},a=434−3r​,
  • If r=30r=30r=30,

a=\frac{34-90}{4}=-14<0,$$ valid.

  • If r=103r=\frac{10}{3}r=310​,

a=\frac{34-10}{4}=6>0,$$ not valid since a<0a<0a<0.

Hence the required radius is

30.\boxed{30}.30​.
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