Question:
The combined equation of the tangent and normal to the curve \( xy = 15 \) at the point (5, 3) is
The combined equation of the tangent and normal to the curve \( xy = 15 \) at the point (5, 3) is
Show Hint
Joint equation of two lines $L_1$ and $L_2$ is found by $L_1 \cdot L_2 = 0$.
Updated On: Apr 30, 2026
- \( 15x^2 - 15y^2 + 16xy = 480 \)
- \( 15x^2 + 16xy - 198x + 10y + 480 - 15y^2 = 0 \)
- \( 15x^2 - 16xy + 19x - 10y - 480 + 15y^2 = 0 \)
- \( 3x + 5y - 30 = 0 \) (simplified individual)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Slopes
$dy/dx = -y/x = -3/5$.
Tangent slope $m_t = -3/5$. Normal slope $m_n = 5/3$.
Step 2: Line Equations
Tangent: $(y-3) = -3/5(x-5) \implies 3x + 5y - 30 = 0$.
Normal: $(y-3) = 5/3(x-5) \implies 5x - 3y - 16 = 0$.
Step 3: Combine
$(3x + 5y - 30)(5x - 3y - 16) = 0$.
$15x^2 - 9xy - 48x + 25xy - 15y^2 - 80y - 150x + 90y + 480 = 0$.
$15x^2 + 16xy - 15y^2 - 198x + 10y + 480 = 0$.
Step 4: Conclusion
Matches Option (B).
Final Answer:(B)
$dy/dx = -y/x = -3/5$.
Tangent slope $m_t = -3/5$. Normal slope $m_n = 5/3$.
Step 2: Line Equations
Tangent: $(y-3) = -3/5(x-5) \implies 3x + 5y - 30 = 0$.
Normal: $(y-3) = 5/3(x-5) \implies 5x - 3y - 16 = 0$.
Step 3: Combine
$(3x + 5y - 30)(5x - 3y - 16) = 0$.
$15x^2 - 9xy - 48x + 25xy - 15y^2 - 80y - 150x + 90y + 480 = 0$.
$15x^2 + 16xy - 15y^2 - 198x + 10y + 480 = 0$.
Step 4: Conclusion
Matches Option (B).
Final Answer:(B)
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Applications of Derivatives Questions
- A wire of length 20 units is divided into two parts such that the product of one part and the cube of the other part is maximum. The product of these parts is:
- 20 meters of wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible surface area, then the radius of the circle is
- The equation of the normal to the curve \( 2x^2 + 3y^2 - 5 = 0 \) at \( P(1, 1) \) is
- A particle moves according to the law \( s = t^3 - 6t^2 + 9t + 25 \). The displacement of the particle at the time when its acceleration is zero is
- If the population grows at the rate of \( 8% \) per year, then the time taken for the population to be doubled is \([ \log 2 = 0.6912 ]\)
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions