Skip to content
Exercise 7.1 · Q21

Q.Find the equation of tangent to the parabola y2=12xy^2 = 12x from the point (2,5)(2,5).

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
14% · 21/151 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

y2=12x⇒4a=12⇒a=3y^2=12x \Rightarrow 4a=12 \Rightarrow a=3. Tangent with slope mm: y=mx+3my=mx+\dfrac{3}{m}.

Passes through (2,5)(2,5): 5=2m+3m⇒5m=2m2+3⇒2m2−5m+3=0⇒(2m−3)(m−1)=05=2m+\dfrac{3}{m} \Rightarrow 5m=2m^2+3 \Rightarrow 2m^2-5m+3=0 \Rightarrow (2m-3)(m-1)=0.

So m=32m=\dfrac32 or m=1m=1.

Using point-slope form through (2,5)(2,5): …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.