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Hyperbola question

2010 · Shift 1 · Q36
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Hyperbola question

2010 · Shift 1 · Q36

JEE AdvancedMathematicsHyperbolaMCQ+4 / −1
The circle x2+y2−8x=0{x^2} + {y^2} - 8x = 0x2+y2−8x=0 and hyperbola x29−y24=1{{{x^2}} \over 9} - {{{y^2}} \over 4} = 19x2​−4y2​=1 intersect at the points AAA and BBB. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
  1. A
    2x−5y−20=02x - \sqrt {5y} - 20 = 02x−5y​−20=0
  2. B
    2x−5y+4=02x - \sqrt {5y} + 4 = 02x−5y​+4=0
  3. C
    3x−4y+8=03x - 4y + 8 = 03x−4y+8=0
  4. D
    4x−3y+4=04x - 3y + 4 = 04x−3y+4=0
View written solutionFree

Correct answer: B

  1. Interpret the given equations

The circle is x2+y2−8x=0x^2+y^2-8x=0x2+y2−8x=0 which can be rewritten as x2−8x+y2=0x^2-8x+y^2=0x2−8x+y2=0 (x−4)2+y2=16(x-4)^2+y^2=16(x−4)2+y2=16 So the circle has center C(4,0)C(4,0)C(4,0) and radius r=4r=4r=4.

The hyperbola is x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=19x2​−4y2​=1 which is a standard hyperbola centered at the origin.

We need a common tangent with positive slope to both the circle and the hyperbola.

  1. Find tangent to the hyperbola in slope form

For the hyperbola x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,a2x2​−b2y2​=1, a tangent with slope mmm is y=mx±a2m2−b2.y=mx\pm \sqrt{a^2m^2-b^2}.y=mx±a2m2−b2​.

Here a2=9a^2=9a2=9, b2=4b^2=4b2=4, so tangent(s) are y=mx±9m2−4.y=mx\pm \sqrt{9m^2-4}.y=mx±9m2−4​.

  1. Condition for the same line to be tangent to the circle

Write the line as y=mx+cy=mx+cy=mx+c or mx−y+c=0.mx-y+c=0.mx−y+c=0.

For this line to be tangent to the circle (x−4)2+y2=16(x-4)^2+y^2=16(x−4)2+y2=16, the perpendicular distance from center (4,0)(4,0)(4,0) to the line must equal radius 444.

So, ∣4m+c∣m2+1=4.\frac{|4m+c|}{\sqrt{m^2+1}}=4.m2+1​∣4m+c∣​=4. Squaring, (4m+c)2=16(m2+1).(4m+c)^2=16(m^2+1).(4m+c)2=16(m2+1).

Now for a tangent to the hyperbola, we have c=±9m2−4.c=\pm \sqrt{9m^2-4}.c=±9m2−4​. Substitute into the circle condition.

  1. Solve for the slope

Take c=±9m2−4.c=\pm \sqrt{9m^2-4}.c=±9m2−4​. Then (4m+c)2=16m2+16.(4m+c)^2=16m^2+16.(4m+c)2=16m2+16. Expanding, 16m2+8mc+c2=16m2+1616m^2+8mc+c^2=16m^2+1616m2+8mc+c2=16m2+16 8mc+c2=16.8mc+c^2=16.8mc+c2=16. But c2=9m2−4,c^2=9m^2-4,c2=9m2−4, so 8mc+9m2−4=168mc+9m^2-4=168mc+9m2−4=16 8mc+9m2=20.8mc+9m^2=20.8mc+9m2=20.

Now test the options instead of solving abstractly.

  1. Check the given options

Option A: 2x−5 y−20=02x-\sqrt5\,y-20=02x−5​y−20=0

Rewrite: y=25x−205y=\frac{2}{\sqrt5}x-\frac{20}{\sqrt5}y=5​2​x−5​20​ So slope m=25>0,c=−205=−45.m=\frac{2}{\sqrt5}>0, \quad c=-\frac{20}{\sqrt5}=-4\sqrt5.m=5​2​>0,c=−5​20​=−45​.

For hyperbola tangent, need c=±9m2−4.c=\pm\sqrt{9m^2-4}.c=±9m2−4​. Now m2=45m^2=\frac{4}{5}m2=54​ 9m2−4=365−4=1659m^2-4=\frac{36}{5}-4=\frac{16}{5}9m2−4=536​−4=516​ 9m2−4=45\sqrt{9m^2-4}=\frac{4}{\sqrt5}9m2−4​=5​4​ But here c=−45≠±45c=-4\sqrt5\neq \pm \frac{4}{\sqrt5}c=−45​=±5​4​, so not tangent to hyperbola.

Hence A is false.

Option B: 2x−5 y+4=02x-\sqrt5\,y+4=02x−5​y+4=0

Rewrite: y=25x+45y=\frac{2}{\sqrt5}x+\frac{4}{\sqrt5}y=5​2​x+5​4​ So m=25,c=45.m=\frac{2}{\sqrt5}, \quad c=\frac{4}{\sqrt5}.m=5​2​,c=5​4​.

Check hyperbola condition: m2=45m^2=\frac{4}{5}m2=54​ 9m2−4=365−4=1659m^2-4=\frac{36}{5}-4=\frac{16}{5}9m2−4=536​−4=516​ 9m2−4=45=c\sqrt{9m^2-4}=\frac{4}{\sqrt5}=c9m2−4​=5​4​=c So this is tangent to the hyperbola.

Now check circle tangency. Distance from (4,0)(4,0)(4,0) to line 2x−5y+4=02x-\sqrt5 y+4=02x−5​y+4=0 is ∣2⋅4+4∣22+(5)2=123=4.\frac{|2\cdot 4+4|}{\sqrt{2^2+(\sqrt5)^2}}=\frac{12}{3}=4.22+(5​)2​∣2⋅4+4∣​=312​=4. So it is also tangent to the circle.

Thus B is correct.

Option C: 3x−4y+8=03x-4y+8=03x−4y+8=0

Rewrite: y=34x+2y=\frac{3}{4}x+2y=43​x+2 So m=34,c=2.m=\frac34, \quad c=2.m=43​,c=2. For hyperbola tangency,

\sqrt{\frac{81}{16}-\frac{64}{16}}=\frac{\sqrt{17}}{4}\neq 2.$$ So not tangent to hyperbola. Hence C is false. --- ### Option D: $4x-3y+4=0$ Rewrite: $$y=\frac43 x+\frac43$$ So $$m=\frac43, \quad c=\frac43.$$ For hyperbola tangency, $$\sqrt{9m^2-4}=\sqrt{9\cdot \frac{16}{9}-4}= \sqrt{16-4}=2\sqrt3\neq \frac43.$$ So not tangent to hyperbola. Hence D is false. --- 6. **Final answer** The required common tangent with positive slope is $$\boxed{2x-\sqrt5\,y+4=0}$$ which is **Option B**.
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