OCR AS Pure 2017 Specimen — Question 8 10 marks

Exam BoardOCR
ModuleAS Pure (AS Pure Mathematics)
Year2017
SessionSpecimen
Marks10
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyStandard +0.3 This is a multi-step problem involving differentiation, normals, and circle equations, but each step follows standard AS-level procedures. Finding k requires equating the normal gradient to the given line's gradient (routine), then using the normal equation to find Q, and finally applying the circle equation formula. While it has multiple parts worth 10 marks, it requires no novel insight—just careful application of standard techniques in sequence, making it slightly easier than average.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

A curve has equation \(y = kx^{\frac{1}{2}}\) where \(k\) is a constant. The point \(P\) on the curve has \(x\)-coordinate 4. The normal to the curve at \(P\) is parallel to the line \(2x + 3y = 0\) and meets the \(x\)-axis at the point \(Q\). The line \(PQ\) is the radius of a circle centre \(P\). Show that \(k = \frac{1}{2}\). Find the equation of the circle. [10]

A curve has equation $y = kx^{\frac{1}{2}}$ where $k$ is a constant.
The point $P$ on the curve has $x$-coordinate 4.
The normal to the curve at $P$ is parallel to the line $2x + 3y = 0$ and meets the $x$-axis at the point $Q$.
The line $PQ$ is the radius of a circle centre $P$.

Show that $k = \frac{1}{2}$.
Find the equation of the circle. [10]

\hfill \mbox{\textit{OCR AS Pure 2017 Q8 [10]}}