The Tangents to the Curve y = (x−2)² − 1 at Its Points of Intersection With the Line x − y = 3 Intersect at the Point

JEE Main 2019. Previous Year Question.

Problem

The tangents to the curve $y=(x-2)^2-1$ at its points of intersection with the line $x-y=3$, intersect at the point:

Key insight. The line meets the curve at exactly two points, and the tangent drawn at each of those points is a completely ordinary tangent line — there’s no shortcut needed beyond finding the intersection points, computing the slope of the curve at each, and solving the resulting pair of straight-line equations.

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Approach

Substitute the line’s equation into the curve’s equation to find the two points of intersection. Differentiate the curve to get a formula for the slope at any point, and evaluate it at both intersection points. Write the tangent line at each point using its slope, and solve the two tangent equations simultaneously to find where they cross.

Solution

Step 1 — Rewrite the line and substitute

From $x-y=3$: $y=x-3$. Substitute into $y=(x-2)^2-1$:

$$x-3 = (x-2)^2-1 = x^2-4x+3$$

$$x^2-5x+6=0 \implies (x-2)(x-3)=0 \implies x=2 \text{ or } x=3$$

Step 2 — Find the two points of intersection

At $x=2$: $y=2-3=-1$, giving point $(2,-1)$. At $x=3$: $y=3-3=0$, giving point $(3,0)$.

Step 3 — Find the slope function

$$\frac{dy}{dx} = 2(x-2)$$

Step 4 — Evaluate the slope at each point

At $(2,-1)$: slope $=2(2-2)=0$. At $(3,0)$: slope $=2(3-2)=2$.

Step 5 — Write the tangent line at each point

At $(2,-1)$ with slope $0$: $\;y-(-1)=0\cdot(x-2) \implies y=-1$.

At $(3,0)$ with slope $2$: $\;y-0=2(x-3) \implies y=2x-6$.

Step 6 — Solve the two tangent equations together

Substituting $y=-1$ into $y=2x-6$:

$$-1=2x-6 \implies 2x=5 \implies x=\frac52$$

Answer

Common mistakes

• Confusing the tangent to the curve at the intersection points with a “chord of contact” without first checking the setup applies. For a general conic, a chord-of-contact shortcut exists, but working through the two tangent lines directly (as done here) is more transparent and less error-prone for a parabola shifted this way.
• Mixing up which point gives which slope. Since $(2,-1)$ gives a horizontal tangent (slope $0$) while $(3,0)$ gives a slanted one (slope $2$), swapping them would produce two very different — and wrong — tangent-line equations.

Practise next

• Find where the tangents to $y=(x-1)^2-4$ at its points of intersection with the line $x-y=5$ meet, using the same substitute-differentiate-and-solve approach.