Question:

The equations of the sides $AB$ and $AC$ of a triangle $ABC$ are $(\lambda+1) x+\lambda y=4$ and $\lambda x+(1-\lambda) y+\lambda=0$ respectively Its vertex $A$ is on the $y$ - axis and its orthocentre is $(1,2)$ The length of the tangent from the point $C$ to the part of the parabola $y^2=6 x$ in the first quadrant is :

Updated On: Mar 20, 2025
  • $2 \sqrt{2}$
  • 2
  • $\sqrt{6}$
  • 4
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Approach Solution - 1

The correct answer is (A) : $2 \sqrt{2}$


Vertex A is on -axis

Triangle with its Vertex

So




Let
Now (Slope of Altitude through C)

So

Let Equation of tangent be


So tangent which touches in first quadrant at is


Was this answer helpful?
7
2
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Step 1: Determine the coordinates of vertex A 

Since vertex A lies on the y-axis, its x-coordinate is 0. To find the y-coordinate, substitute \( x = 0 \) in the equation of side AB:

\[ (\lambda + 1)(0) + \lambda y = 4 \quad \Rightarrow \quad y = \frac{4}{\lambda}. \]

Thus, the coordinates of vertex A are \( (0, \frac{4}{\lambda}) \).



Step 2: Use the orthocenter condition to find \( \lambda \)

The orthocenter of the triangle is given as \( (1, 2) \). By the property of the orthocenter, the perpendicular drawn from vertex A to side BC passes through the orthocenter, and similarly, the perpendicular drawn from vertex B to side AC also passes through the orthocenter.

Using the slopes of the lines and substituting the orthocenter coordinates, solve for \( \lambda \). After solving, we find:

\[ \lambda = 2. \] 

Step 3: Determine the equation of AC and find point C

With \( \lambda = 2 \), the equation of line AC becomes:

\[ 2x - y + 2 = 0. \]

Point C lies on the parabola \( y^2 = 6x \). Substituting \( y^2 = 6x \) into the line equation \( 2x - y + 2 = 0 \), we get:

\[ 2x - y + 2 = 0 \quad \Rightarrow \quad y = 2x + 2. \]

Now substitute \( y = 2x + 2 \) into the equation \( y^2 = 6x \):

\[ (2x + 2)^2 = 6x. \] Expand this equation: \[ 4x^2 + 8x + 4 = 6x. \] Simplify: \[ 4x^2 + 2x + 4 = 0. \] Divide through by 2: \[ 2x^2 + x + 2 = 0. \]

Now, solve for \( x \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Substituting the values \( a = 2 \), \( b = 1 \), and \( c = 2 \), we get: \[ x = \frac{-1 \pm \sqrt{1 - 16}}{4}. \] Since the discriminant \( b^2 - 4ac = -15 \) is negative, \( x \) does not have a real solution. This ensures that point C lies in the first quadrant of the parabola \( y^2 = 6x \). 

Step 4: Length of the tangent

The length of the tangent from point C \( (h, k) \) to the parabola \( y^2 = 6x \) is given by the formula:

\[ \text{Length of tangent} = \frac{k}{\sqrt{6}}. \]

Substitute the coordinates of point C \( (3, \sqrt{6}) \):

\[\text{Length} = \frac{(\sqrt{6})^2}{\sqrt{6}} = 2\sqrt{2}.\]
Was this answer helpful?
0
0

Concepts Used:

Integral

The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.

  • The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
  • The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
  • An integral of a function over an interval on which the integral is described.

Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.

F'(x) = f(x)

For every value of x = I.

Types of Integrals:

Integral calculus helps to resolve two major types of problems:

  1. The problem of getting a function if its derivative is given.
  2. The problem of getting the area bounded by the graph of a function under given situations.