Question

The \(x\)-intercept of one of the common tangents to the circle \[ 3x^2+3y^2=169 \] and the parabola \[ y^2=26x \] is:

• A. \(\dfrac{13}{2}\)
• B. \(-13\)
• C. \(13\)
• D. \(-\dfrac{13}{2}\)

Hint:

For the parabola \(y^2=4ax\), always remember the slope form of tangent: \[ y=mx+\frac{a}{m}. \] It is extremely useful when dealing with common tangent problems.

Answer 1

The Correct Option is B

Concept: A common tangent must satisfy the tangent condition for both the circle and the parabola simultaneously. For the parabola \[ y^2=4ax, \] the tangent of slope \(m\) is \[ y=mx+\frac{a}{m}. \] We first obtain the tangent equation from the parabola and then impose the condition that the same line is tangent to the circle.

Step 1: Rewrite the parabola in standard form. Given \[ y^2=26x. \] Comparing with \[ y^2=4ax, \] we get \[ 4a=26 \] \[ a=\frac{13}{2}. \] Therefore the tangent of slope \(m\) is \[ y=mx+\frac{13}{2m}. \]

Step 2: Rewrite the circle. Given \[ 3x^2+3y^2=169. \] Dividing by \(3\), \[ x^2+y^2=\frac{169}{3}. \] Hence \[ r=\frac{13}{\sqrt3}. \]

Step 3: Use tangency condition for the circle. The tangent is \[ mx-y+\frac{13}{2m}=0. \] Distance from the centre \((0,0)\) to this line must equal the radius. \[ \frac{\left|\frac{13}{2m}\right|} {\sqrt{m^2+1}} = \frac{13}{\sqrt3}. \] Cancelling \(13\), \[ \frac1{2|m|\sqrt{m^2+1}} = \frac1{\sqrt3}. \] Squaring, \[ 4m^2(m^2+1)=3. \] \[ 4m^4+4m^2-3=0. \] Let \[ t=m^2. \] Then \[ 4t^2+4t-3=0. \] \[ (2t+3)(2t-1)=0. \] \[ t=\frac12. \] Thus \[ m=\pm \frac1{\sqrt2}. \]

Step 4: Find the tangent having negative \(x\)-intercept. Using \[ m=\frac1{\sqrt2}, \] the tangent is \[ y=\frac{x}{\sqrt2}+\frac{13\sqrt2}{2}. \] For \(x\)-intercept, \[ y=0. \] Hence \[ 0=\frac{x}{\sqrt2}+\frac{13\sqrt2}{2}. \] Multiplying by \(\sqrt2\), \[ x+13=0. \] \[ x=-13. \]

Step 5: Final Answer. Therefore, the required \(x\)-intercept is \[ \boxed{-13}. \]