How many points of intersection are there between the parabola \(y=x^2-2kx+k^2+9\) and the \(x\)-axis?
Answer and explanation
Correct answer: No intersection
On the \(x\)-axis, \(y=0\). Therefore, the intersection equation is \(x^2-2kx+k^2+9=0\), or \((x-k)^2+9=0\). Since \((x-k)^2\geq 0\), the left side is always at least 9 and can never be zero. Equivalently, the discriminant is \(D=(-2k)^2-4(k^2+9)=-36<0\), so there are no real points of intersection. The tempting answers of one or two intersections are incorrect because they would require a zero or positive discriminant. Exam tip: completing the square gives the result immediately.
Frequently asked questions
What is the correct answer to this question?
No intersection
Why is this the correct answer?
On the \(x\)-axis, \(y=0\). Therefore, the intersection equation is \(x^2-2kx+k^2+9=0\), or \((x-k)^2+9=0\). Since \((x-k)^2\geq 0\), the left side is always at least 9 and can never be zero. Equivalently, the discriminant is \(D=(-2k)^2-4(k^2+9)=-36<0\), so there are no real points of intersection. The tempting answers of one or two intersections are incorrect because they would require a zero or positive discriminant. Exam tip: completing the square gives the result immediately.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
