Question:

The curves \[ y=x^2-1,\qquad y=8x-x^2-9 \]

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If two curves intersect at a point and their slopes at that point are equal, then the curves touch each other at that point.
Updated On: Jun 26, 2026
  • intersect at right angles at \((2,3)\)
  • touch each other at \((2,3)\)
  • intersect at \(45^\circ\)
  • intersect at \(60^\circ\)
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The Correct Option is B

Solution and Explanation

Step 1: Find the point of intersection.
The given curves are \[ y=x^2-1 \] and \[ y=8x-x^2-9. \] At the point of intersection, \[ x^2-1=8x-x^2-9. \] So, \[ 2x^2-8x+8=0. \] Dividing by \(2\), \[ x^2-4x+4=0. \] Thus, \[ (x-2)^2=0. \] Hence, \[ x=2. \] Substitute \(x=2\) in \[ y=x^2-1. \] \[ y=2^2-1=4-1=3. \] Therefore, the curves intersect at \[ (2,3). \]

Step 2: Find the slopes of both curves at \((2,3)\).
For the first curve, \[ y=x^2-1. \] Differentiating with respect to \(x\), \[ \frac{dy}{dx}=2x. \] At \(x=2\), \[ m_1=2(2)=4. \] For the second curve, \[ y=8x-x^2-9. \] Differentiating with respect to \(x\), \[ \frac{dy}{dx}=8-2x. \] At \(x=2\), \[ m_2=8-2(2)=4. \]

Step 3: Compare the slopes.
Since \[ m_1=m_2=4, \] the two curves have the same tangent at the point \((2,3)\).
Therefore, the curves touch each other at \((2,3)\).

Step 4: Final conclusion.
Hence, \[ \boxed{\text{touch each other at }(2,3)} \]
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