Question

Parabola \( y = x^2 + px + q \) is passing through \( (1,-1) \) and vertex of parabola is at minimum distance from x-axis then \( p^2 + q^2 \) is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For a parabola \( y=ax^2+bx+c \), the vertex y-coordinate is found by substituting \( x=-\frac{b}{2a} \). If a minimum distance from the x-axis is asked, minimize the absolute value of this y-coordinate using the given condition.

Answer 1

The Correct Option is D

Step 1: Use the condition that the parabola passes through \( (1,-1) \).
Given parabola is:
\[ y=x^2+px+q \] Since it passes through the point \( (1,-1) \), substitute \( x=1 \) and \( y=-1 \):
\[ -1=1+p+q \] \[ p+q=-2 \] So, we get:
\[ q=-2-p \]
Step 2: Find the y-coordinate of the vertex.
For the parabola \( y=x^2+px+q \), the x-coordinate of the vertex is:
\[ x_v=-\frac{p}{2} \] Now substitute this into the parabola to get the y-coordinate of the vertex:
\[ y_v=\left(-\frac{p}{2}\right)^2+p\left(-\frac{p}{2}\right)+q \] \[ y_v=\frac{p^2}{4}-\frac{p^2}{2}+q \] \[ y_v=q-\frac{p^2}{4} \] Using \( q=-2-p \), we get:
\[ y_v=-2-p-\frac{p^2}{4} \]
Step 3: Use the condition of minimum distance from x-axis.
The distance of the vertex from the x-axis is \( |y_v| \). For minimum distance, this value must be minimum. Since the minimum possible value of an absolute quantity is 0, we set:
\[ y_v=0 \] Thus:
\[ -2-p-\frac{p^2}{4}=0 \] Multiply by 4 throughout:
\[ -8-4p-p^2=0 \] \[ p^2+4p+8=0 \] Actually, rewrite carefully from the vertex form condition using the parabola relation:
\[ y_v=q-\frac{p^2}{4} \] and for minimum distance from x-axis, vertex lies on x-axis, so:
\[ q-\frac{p^2}{4}=0 \] Hence:
\[ q=\frac{p^2}{4} \] Now using \( p+q=-2 \):
\[ p+\frac{p^2}{4}=-2 \] Multiply by 4:
\[ p^2+4p+8=0 \] This gives no real value, so now use the idea that the vertex distance is minimum under the constraint \( q=-2-p \). Define:
\[ f(p)=q-\frac{p^2}{4}=-2-p-\frac{p^2}{4} \] Distance is minimum when the vertex is nearest to x-axis, so the quadratic expression must attain its greatest possible value because it is always negative. Differentiate:
\[ f'(p)=-1-\frac{p}{2} \] Set \( f'(p)=0 \):
\[ -1-\frac{p}{2}=0 \] \[ p=-2 \] Then:
\[ q=-2-(-2)=0 \]
Step 4: Calculate \( p^2+q^2 \).
Now:
\[ p=-2,\qquad q=0 \] Therefore:
\[ p^2+q^2=(-2)^2+0^2 \] \[ =4 \]
Step 5: Conclusion.
Hence, the required value of \( p^2+q^2 \) is \( 4 \).
Final Answer: \( 4 \)