Question

If $m_1$ and $m_2$ are the slopes of the tangents drawn from the point (1,4) to the parabola $y^2 = 11x$ then $2(m_1^2 + m_2^2) =$

• A. 24
• B. 22
• C. 21
• D. 18

Hint:

The equation of a tangent in slope form ($y=mx+a/m$) is very useful for problems involving tangents from an external point to a parabola. Substituting the point's coordinates leads to a quadratic equation whose roots are the slopes of the tangents.

Answer 1

The Correct Option is C

The equation of a tangent to the parabola $y^2=4ax$ in terms of slope $m$ is $y = mx + \frac{a}{m}$.
For the given parabola $y^2 = 11x$, we have $4a=11$, so $a=11/4$.
The equation of the tangent is $y = mx + \frac{11}{4m}$.
This tangent passes through the point $(x_1, y_1) = (1,4)$. So, we substitute these coordinates into the equation.
$4 = m(1) + \frac{11}{4m}$.
$4 = m + \frac{11}{4m}$.
Multiply the entire equation by $4m$ to clear the denominator:
$16m = 4m^2 + 11$.
Rearrange into a standard quadratic form in $m$:
$4m^2 - 16m + 11 = 0$.
The roots of this quadratic equation are the slopes $m_1$ and $m_2$.
Using Vieta's formulas:
Sum of slopes: $m_1 + m_2 = -\frac{-16}{4} = 4$.
Product of slopes: $m_1 m_2 = \frac{11}{4}$.
We need to find the value of $2(m_1^2 + m_2^2)$.
We can express $m_1^2 + m_2^2$ in terms of the sum and product:
$m_1^2 + m_2^2 = (m_1+m_2)^2 - 2m_1m_2$.
$m_1^2 + m_2^2 = (4)^2 - 2(\frac{11}{4}) = 16 - \frac{11}{2} = \frac{32-11}{2} = \frac{21}{2}$.
Now, we calculate the final expression:
$2(m_1^2 + m_2^2) = 2 \left(\frac{21}{2}\right) = 21$.