At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$. Find the equation of the curve given that it passes through $(-2,1)$.

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Calculus 2 / BC

NCERT Class 12 Part 2 - Math

Chapter 9

Differential Equations

Section 4

Formation of a Differential Equation whose General Solution is given

Discussion

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Video Transcript

we will not find the equation of Ako which passes through the point minus to come over. And we are also given this information. That is a slope of the tangent is twice the line segment turning the point X comma y to the point minus four. Come up minus three. So basically we are going to form the differential equation first using the given uh conditions. Then we sold the differential equation. That is we find the general solution of the differential equation and finally we find the particular solution using this condition which in fact will be the equation of the co. So let's go ahead and form the differential equation first. Using this information. We are given slope of tangents so we can write down this as dy by dx and this is equal to twice the slope of the line segment joining the point. There is a slope is missing here so that I don't hear uh joining the point X comma Y to the point minus four Command ministry. So first let find the slope of the line joining these two points for this will be placing this slow formula that is Y tu minus Y one over X two minus X one. So this we can take it as X one, Y 1 and thus we take it as X two, Y two. Which means this will be why- of -3. Well what X- of -4 which in fact is white plus three. Can I don't This is Y plus three divided by express for No. So it's given us twice the slope of the line segment. So this is the slope of the line segment joining these two points. So this slope of the tangent to the co the CBD excess twice this look. So we can write on this as Y plus three Over X-plus four. Yeah. So we are done with the first step that is we have formed a differential equation. Now let's go ahead and solve it. That is we find the general solution for this. I'm going to move this Y plus three to this side that is to the left side and the eggs to the and hindsight. So which means uh had to divide by Y. Place three first and simultaneously I had to multiply by uh dx So when I do that be getting uh do you worry DY by Y Plus three. We'll have this site. We have this too and this will be the X. Boy, X plus four. So now we have successfully separated the variable cities all the way terms on the left side, that's place all the X terms on the right side. Now we can integrate. So let's do that. We can put this integration that is this too is pulled out of the integration. So we can use the standard formula of for this integration that is A D. Y divided by Y plus three. When we integrate we we will be getting log off. Y plus three on the website. Similarly on the right side we can use the same formula that is dx over express board is Lago X plus for and then we will put the integration constant this time I'm going to put this as log off the so that I can combine these two terms. So in the next step is in place this side remains the same as a log of y plus three is equal to using the property of logarithms. This will get place to the power of X plus three. So I can write down this this log off X plus four. Place to the power of two plus log see So now let's simplify the right side so we can apply the property of low growth and that is so this is a log of a press log of B is log off maybe. So I can combine it as log off A plus B. That is I put the C in front see time's up X plus four square everything. And at this particular law algorithm. So this side we still have this uh log off white plus three. Now we can remove the algorithm on both sides which means this side will be getting oil plus three. Is he called to see to himself? X plus four quantity squared. So this in fact is there general solution of the differential equation that we just form now we have to find the particular solution. Otherwise we have to find the equation of the so for that I will be placing this information that is the co passes through the point minus two comma one. So let's plug this point into this equation. So this is a general solution and we played the point minus two comma one. This helps us to find the value of C into this equation. So we plug uh Y equal to one, which means this will be one plus three equal to four. Uh And similarly this is see time's up Access -2. So therefore -2 plus forests 2, 2 squared is four. So this is we have 4C equal to four. So dividing by foreign voices, we get the value of c equal to one and we know substitute the value of C into the situation which will help us to find the particular solution or otherwise. Which in fact is the equation of the code. Maybe I could put this in bracket, We are finding the equation of the curve, which is basically a particular solution. So this is equal to y plus three is equal to replace this. Uh See by one. So when we do that we'll be getting one times of express for all square. So we'll be getting y plus three is equal to explainable. All squared. So this is the equation of the code

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