Find a point on the curve x^2+2 y^2=6 whose distance from the line x+y=7 is minimum.

Name: Problem 52, Chapter 9: Differential Calculus Booster with Problems and Solutions for JEE Main and Advanced
Uploaded: 2022-04-15T17:54:43-08:00
Duration: 3 min
Channel: Monisha Shukla
Description: Find a point on the curve x^2+2 y^2=6 whose distance from the line x+y=7 is minimum.

step 1 Hello, our question is find the curve. So first given curve is here. First given curve is here x

Find a point on the curve x^2+2 y^2=6 whose distance from...

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. Understanding their algebraic representations, such as quadratic equations, is crucial in analyzing and solving problems involving these curves, particularly when optimization problems require working within the constraints of a conic section.

Constrained Optimization

This concept involves finding the maximum or minimum of a function subject to one or more constraints. It is widely used in various fields of mathematics and science to analyze systems where the solution must adhere to certain conditions, employing techniques that ensure the optimum is found while satisfying the given restrictions.

Lagrange Multipliers

Lagrange multipliers are a method used to solve constrained optimization problems. By setting the gradient of the function equal to a constant times the gradient of the constraint function, one can determine critical points where the function could achieve a maximum or minimum under the given constraint. This technique simplifies finding extrema without having to explicitly solve the constraint for one variable.

Distance from a Point to a Line

This concept involves calculating the shortest distance between a given point and a specified line using the distance formula. In a Cartesian plane, the distance from a point to a line is determined by the absolute value of the line’s equation evaluated at the point, divided by the square root of the sum of the squares of the coefficients of the variables. This formula is fundamental in geometry and optimization problems where minimizing or maximizing distances is essential.

Transcript

00:01 Hello, our question is find the curve.

00:01 So first given curve is here.

00:03 First given curve is here x square plus 2 y square which is equals to 6.

00:07 Whose distance with the line and given line is here, x plus y equals to 7 is minimum.

00:13 We have to find that point from this curve to this line.

00:18 The distance is shortest here.

00:20 So as we can see here, our curve is x square plus our curve is here.

00:24 X square plus 2 y square equals to 6.

00:26 So this equation, by simplifying this equation, we can write this.

00:30 Has x square by 6 plus y square by 3 which is equals to 1 which is equation of here ellipse this one is equation of an ellipse so and this one is equation of straight line this equation is of straight line here so the line so the tangent seems the tangent is parallel to the line here as we can see this one is equation of ellipse this one is equation of straight line so this straight line is parallel to the tangent here so because here d y by d x d .y by d x.

01:01 D .x.

01:01 D differentiate both sides with respect to x, we can write it as here one plus d .y by dx equals to 0.

01:06 This one is implying d .y by dx for the straight line here, this one equals to minus of 1.

01:12 So by this equation here, if we differentiate this ellipse, we can write it as here.

01:18 A differentiation, this one is equation first.

01:20 Differentiate equation first.

01:24 With respect to x both side, we can write it as here 2x divided by 6 plus 2y divided by 3.

01:30 D .y by d x is equals to 0 here or we can write it as here x by 3 minus equals to here y by 3 d.

01:39 D .x so d...

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